Class 12 Engineering Physics - Electric Charges And Fields

Find the excess (equal in number) of electrons that must be placed on each of two small spheres spaced 3 cm apart with a force of repulsion between the spheres to be $10^{-19}N$

Force between the two electrons,

$F=\dfrac {1}{4\pi \epsilon_0}\times \dfrac {(ne)(ne)}{d^2}$
or

$n^2=\dfrac {F\times 4\pi \epsilon_0\times d^2}{e^2}$

$\displaystyle n^2=\dfrac {10^{-19}\times \dfrac {1}{9\times 10^9}\times (3\times 10^{-2})^2}{(1.6\times 10^{-19})^2}$

$=(625)^2$
or

$n=625$

$\therefore$ The number of electrons placed on each of the sphere

$=625.$

The electric force on a charge $q_1$ due to $q_2$ is $(4\widehat{i} - 3\widehat{j})$ N. What is the force on $q_2$ due to $q_1$ ?

The force on

$q_2$ due to

$q_1$ will be equal and opposite to force on

$q_1$ due to

$q_2$ . (By Newton's third law)

Thus, the force on

$q_2$ due to

$q_1$ will be

$[-(4\hat i-3\hat j)] N$

Does the force on a charge due to another charge depend on the charges present nearby?

Force on a charge due to another charge =

$\dfrac{1}{4 \pi \epsilon_0}{q_1 q_2\times r^2}$
This force does not depend on other charges

Is there any lower limit to the electric force between two particles placed at a certain distance?

Yes. The electric force between two charged is given by:

$F=\frac{kq_1 q_2}{r^2}$
Now when

$r \rightarrow \infty$ ,

$F=0$ which is the minimum value of electric force.
But if two charges are 'd' distance apart then also minimum value of force can be calculated using the fact that electric charge is quantized and it has a minimum value of

$1.6 \times 10^{-19} C$ .

Find the coulombic force between two $\alpha$ - particles separated by a distance of 8.2 fermi, in air.

Charge of each

$\alpha - particle,$

$q=2e$

$=2\times 1.6\times 10^{-19}$

$=3.2\times 10^{-19}C$
Distance between the

$\alpha-particles, d=8.2 fermi=8.2\times 10^{-15}m$

$\therefore$ Coulomb force between the two

$\alpha-particles, F=\dfrac {1}{4\pi \epsilon_0}\dfrac {q_1q_2}{d^2}$

$\displaystyle =\frac {9\times 10^9\times (3.2\times 10^{-19})(3.2\times 10^{-19})}{(8.2\times 10^{-15})^2}$

$\approx 13.7 \approx 14N$

Two charged particles having charge $2\times 10^{-8} C$ each are joined by an insulating string of length 100 cm. The system is kept on a horizontal table. Find the tension in the string.

Here tension in the string will be equal to the
So

$T=F=\dfrac {1}{4\pi \epsilon_0}\dfrac {q_1q_2}{r^2}$

$=9\times 10^9\times \dfrac {(2\times 10^{-8})(2\times 10^{-8})}{(1)^2}$

$=3.6\times 10^{-6}N$

Two point charges $q_1$ and $q_2$ are 3 m apart and their combined charge is $20\mu C$ . If one repels the other with a force of 0.075 N, what are two charges?

Let

$q_1q_2$ be two charges
Given

$q_1+q_2=20\times 10^{-6}C$
From Coulomb's law,

$F=\dfrac {1}{4\pi \epsilon_0}\dfrac {q_1q_2}{r^2}$

$\Rightarrow 0.75=9\times 10^9\times \dfrac {q_1q_2}{(3)^2}$

$\Rightarrow q_1q_2=\dfrac {0.075\times 9}{9\times 10^9}=75\times 10^{-12}C^2$

$(q_1=q_2)^2=(q_1+q_2)^2-4q_1q_2$

$=(20\times 10^{-6})^2-4\times 75\times 10^{-12}$

$=10^{-10}C^2$
or

$q_1-q_2=10^{-5}C=10\times 10^{-6}C$

$(q_1+q_2)+(q_1-q_2)$

$=(20\times 10^{-6})+(10\times 10^{-6})$

$=30\times 10^{-6}$
or

$2q_1=30\times 10^{-6}C$

$\Rightarrow q_1=15\times 10^{-6}C=15\mu C$
then,

$q_2=20\times 10^{-6}-15\times 10^{-6}$

$=5\times 10^{-6}=5\mu C$

Two identical spheres having charges $+q_1$ and $+q_2$ are kept at a distance "d". These two are brought in contact with an insulated handle and again separated to the same distance. Find the force $F_2$ between them in terms of initial force $F_1$ .

Force between the two charges (

$+q_1 and +q_2)$ , if they are at a distance 'd' is

$F_1=\dfrac {1}{4\pi \epsilon_0}\dfrac {q_1q_2}{d^2}.....(1)$
When the metal spheres are in contact the charge flow from one sphere to another till both the spheres acquire the same charge.
Here

$q_1 and q_2$ are of same sign, hence after contact each sphere will have a charge

$\dfrac {(q_1+q_2)}{2}$
The new charge

$Q=\dfrac {q_1+q_2}{2}$
Now the force between them,

$F_2=\dfrac {1}{4\pi \epsilon_0}\dfrac {[\dfrac {q_1+q_2}{2}]^2}{d^2}.....(2)$
From equations (1) and (2)

$F_2=F_1\dfrac {(q_1+q_1)^2}{4q_1q_2}$

Two identical condncting spheres, fixed in space, attract each other with an electrostatic force of 0.108 N when separated by 50.0 cm, center-to-center. A thin conducting wire then connects the spheres, When the wire is removed, the spheres repel each other with an electrostatic force of 0.0360 N. What were the initial charges on the spheres?

initial charges =

$q_1 and -q_2$
Force of attraction =

$0.108 N = \frac{1}{4 \pi \epsilon} \frac{ q_1 q_2}{0.5^2}$
charges after the connection is made and removed =

$\frac{(q_1 - q_2 )}{2} and \frac{(q_1 - q_2 )}{2}$
Force of repulsion =

$0.036 N = \frac{1}{4 \pi \epsilon} \frac{ (q_1- q_2)^2}{4 \times 0.5^2}$
therefore,

$q_1 q_2 = 3.0012 \times 10^{-12}$ and

$(q_1 - q_2 )^2 = 4 \times 10^{-12}$

$\Rightarrow q_1 =q_2 = 1 \mu F$

(a) Two insulated charged copper spheres A and B have their centres separated by a distance of 50 cm. What is the mutual force of electrostatic repulsion if the charge on each is $6.5\times 10^{-7}$ C? The radii of A and B are negligible compared to the distance of separation.
(b) What is the force of repulsion if each sphere is charged double the above amount, and the distance between them is halved?

$\large \textbf{Part (a):}$

$\textbf{Step 1: Electrostatic repulsion force Initially}$

$Q_{1} = Q_{2} = 6.5 \times 10^{-7}\ C , r = 50\ cm = 0.5\ m$

By Coulomb's Law:

$F = \dfrac{K Q_1 Q_2}{r^2} = \dfrac{9 \times 10^{9} \times 6.5 \times 10^{-7}C \times 6.5 \times 10^{-7}C}{(0.5m)^2}$

$F = 1.52 \times 10^{-2}\ N$

$\large \textbf{Part (b):}$

$\textbf{Step 2: Electrostatic force in second case}$

When charges are doubled and distance is halved.

So,

$Q_{1}' = Q_{2}' = 2 \times 6.5 \times 10^{-7} \ C ,\ \ \ \& \ \ \ r' = 25\ cm = 0.25\ m$

$F' = \dfrac{KQ_{1}' Q_{2}'}{(r')^2} = \dfrac{9 \times 10^{9} \times (2 \times 6.5 \times 10^{-7}C)^2}{(0.25m)^2}$

$F' = 0.24\ N$

What is the force between two small charged spheres having charges of $2\times 10^{-7}$ C and $3\times 10^{-7}$ C placed 30 cm apart in air?

Charge on the first sphere, $q_1= 2 \times 10^{-7}$ C

Charge on the second sphere, $q_2= 3 \times 10^{-7}$ C

Distance between the spheres, $r = 30 cm = 0.3 m$

$F=\displaystyle \frac{q_1q_2}{4\pi \varepsilon_0r^2}$

Where, $\varepsilon_0=$ Permittivity of free space

$\displaystyle \frac{1}{4\pi\varepsilon_0}=9\times 10^9\, N\,m^2C^{-2}$

$\displaystyle F=\frac{9\times 10^9\times 2\times 10^{-7}\times 3\times 10^{-7}}{(0.3)^2}=6\times 10^{-3}N$

Hence, force between the two small charged spheres is $6 \times 10^{-3}$ N. The charges are of same nature. Hence, force between them will be repulsive.

The electrostatic force on a small sphere of charge $0.4\mu C$ due to another small sphere of charge $-0.8\mu C$ in air is 0.2 N.
(a) What is the distance between the two spheres?
(b) What is the force on the second sphere due to the first?

(a)

Electrostatic force on the first sphere, $F = 0.2 N$

Charge on this sphere, $q_1=0.4\mu C = 0.4\times 10^{-6}C$

Charge on the second sphere, $q_2=-0.8\mu C=-0.8\times 10^{-6}C$

Electrostatic force between the spheres is given by the relation,

$\displaystyle F=\frac{q_1q_2}{4\pi \varepsilon_0r^2}$

And, $\displaystyle \frac{1}{4\pi \varepsilon_0} = 9\times 10^9N\,m^2C^{-2}$

Where, $\varepsilon_0=$ Permittivity of free space

$r^2=\displaystyle \frac{q_1q_2}{4\pi \varepsilon_0F}$

$\displaystyle =\frac{04\times 10^{-6}\times 8\times 10^{-6}\times 9\times 10^{9}}{0.2}$

$=144\times 10^{-4}$

$r =\sqrt{144\times 10^{-4}}$

$=0.12m$

The distance between the two spheres is 0.12 m.

(b) Equal and Opposite Force acts on the other sphere (By newton's Third Law). Hence 0.2 N attractive.

Exercise:
[
Two insulated charged copper spheres A and B have their centres separated by a distance of 50 cm. The charge on each is $6.5\times 10^{7}$ C? The radii of A and B are negligible compared to the distance of separation.
]
Suppose the spheres A and B in Exercise have identical sizes. A third sphere of the same size but uncharged is brought in contact with the first, then brought in contact with the second, and finally removed from both. What is the new force of repulsion between A and B?

$Q_A = 6.5\times 10^{-7}$

$Q_B = 6.5\times 10^{-7}$
Let the uncharged sphere be C.

$Q_C = 0$
After contacting with A, the charge gets divided equally between A and C.
So, the new charges on A and C are

$6.5\times 10^{-7}/2$ and

$6.5\times 10^{-7}/2$ , respectively.
Now when we brought into contact C and B, the total charge on B and C again gets divided equally between B and C.
So, the new charges on B and C are

$\displaystyle \frac{3.25\times 10^{-7}+6.5\times 10^{-7}}{2}$ and

$\displaystyle \frac{3.25\times 10^{-7}+6.5\times 10^{-7}}{2}$ , respectively.
So now,

$Q_A' = 3.25\times 10^{-7}$

$Q_B' = 4.875\times 10^{-7}$
Electrostatic force now is

$F=Q_A'Q_B'/4\pi\epsilon_or^{2}$

$=5.7\times 10^{-8}\ N$

Calculate the height of the potential barrier for a head on collision of two deuterons.

For a head on collision, distance between centers of two deuterons=

$2\times radius=r=2\times 2 fermi=4\times 10^{-15}m$

Charge of each deuteron=

$1.6\times 10^{-19}C$

Potential energy=

$\dfrac{e^2}{4\pi \epsilon_0r}=360keV$

Potential energy=

$2\times$ Kinetic Energy of Deuteron

$=360keV$
This is the measure of height of Coulomb Barrier Potential.

Thus Kinetic energy of each deuteron=

$180keV$ .

The charge and coordinates of two charged particles held fixed in an $y$ plane are $q_1=+3.0\ \mu C, x_1=3.5\ cm, y_1=0.50\ cm$ , and $q_2 =-4.0\mu C, x_2 =-2.0\ cm, y_2 =1.5\ cm$ . Find
direction of the electrostatic force on particle $2$ due to tude ?

The vector

$\vec F_{21}$ is directed towards

$q_1$ and makes angle

$\theta$ with the

$+x$ axis, where

$\theta =\tan^{-1}\left(\dfrac {y_2- y_1}{x_2 -x_1}\right) =\tan^{-1}\left(\dfrac {1.5\ cm-0.5\ cm}{-2.0\ cm-3.5\ cm}\right)=-10.3^o \approx -10^o$ .

The electrostatic force on a small sphere of charge $0.4\ \mu C$ due to another small sphere of charge $0.8\ \mu C$ in air is $0.2\ N$ .
(a) What is the distance between the two spheres?
(b) What is the force on the second sphere due to the first?

a) Using Coulomb's Law,

$F=\dfrac{1}{4\pi \varepsilon _{0}}\dfrac{q_{1}q_{2}}{r^{2}}$

$r^{2}=9\times 10^{9}\times \dfrac{0.4\times0.8\times10^{-12}}{0.2}$

$r^{2}=14.4\times 10^{-3}$

$=144\times 10^{-4}$

$r=12\times 10^{-2}=0.12\; m$

b) Force on second sphere due to first is

$0.2\; N$ but in opposite direction, because the charges due to same and distance is same; so force is also same.

Two point charges of $2\times 10^{-7}C$ and $1.0\times 10^{-7}$ are $1.0\ cm$ apart. What is the magnitude of the field produced by either charge at the site of the other? Use standard value of $1/4\pi \epsilon_{0}$

Electric field producedby A on B;

$E=\dfrac{1}{4\pi \varepsilon _{0}}\:\dfrac{q}{r^{2}}$

$=9\times 10^{9}\times \dfrac{10^{-7}}{(10^{-2})^{2}}$

$=9\times 10^{6} \: NC^{-1}$

Electric field produced by charge at B on the charge at A is,

$E=9\times 10^{9}\times \dfrac{2\times10^{-7}}{(10^{-2})^{2}}$

$=18\times 10^{6} \: NC^{-1}$

The electrostatic force on a small sphere of charge $0.4 \mu C$ due to another small sphere of charge $-0.8\mu C$ in the air is $0.2\ N$ .
What is the force on the second sphere due to the first?

Given that force on a sphere of charge

$0.4\mu C$ due to another sphere of charge

$-0.8\mu C$ is

$0.2\, N$ . We have to find the force on the second sphere due to first.

Since two charges are:

$q_{1}=0.4\mu C$

$q_{2}=-0.8\mu C$

So, force is attractive and is equal to

$0.2\, N$

And force on

$-0.8\mu C$ charge due to

$0.4\mu C$ charge is also same since the medium of two charges and distance between the charges is kept same.

So, force is also same, equal to

$0.2\, N$ but its direction is opposite.

The electrostatic force on a small sphere of charge $0.4\mu C$ due to another small sphere of charge $-0.8\mu C$ in the air is $0.2\ N$ .
What is the distance between the two spheres?

Using Coulomb's Law,

$F=\dfrac{1}{4\pi \varepsilon _{0}}\:\dfrac{q_{1}q_{2}}{r^{2}}$

$r^{2}=\dfrac{1}{4\pi \varepsilon _{0}}\:\dfrac{q_{1}q_{2}}{F}$

$r^{2}=\dfrac{9\times10^{9}\times 0.4\times (0.8)\times 10^{-12}}{0.2}$

[Here

$q_{1}=0.4 \; \mu C; \:\: q_{2}=-0.8 \; \mu C;\:\: F=0.2 \; N]$

$r^{2}=14.4\times 10^{-3}$

$=144\times 10^{-4}$

So,

$r=12\times 10^{-2}$

$r=12\:cm$

Two equal charges placed in air and separated by a distance of $2\ m$ repel each other with a force of $10^{-4}kgf$ . Calculate the magnitude of either of the charges.

Let each charge be

$q$ .

Using Coulomb's Law,

$F=\dfrac{1}{4\pi \varepsilon _{0}}\:\dfrac{q^{2}}{r^{2}}$

Here,

$r=2\;m$

$F=10^{-4}\times 9.8\:N$

$=10^{-4}\times 10\:N$

$q^{2}=F\times 4\pi \varepsilon _{0}\times r^{2}$

$=10^{-4}\times \dfrac{10\times 1}{9\times 10^{9}}\times (2)^{2}$

$=\dfrac{10^{-12}}{9}\times (2)^{2}$

So,

$q=\dfrac{2}{3}\times 10^{-6}$

$=0.66\;\mu C$

Each charge is

$0.66\; \mu C$

The electrostatic force of repulsion between two equal positively charged ions is $3.7\times 10^{-9}N$ , when they are separated by a distance of $5\overset {\circ}{A}$ . How many electrons are missing from each ion?

Given, the force of repulsion between two equally positively charged ions

$=3.7\times 10^{-19}\, N$

Distance of separation,

$d=5$ Å

$=5\times 10^{10}\, m$

To find the number of electrons missing from each ion.

Let us suppose charge is

$q_{1}=q_{2}=q$

Let

$n$ be the number of electrons missing.

Using Coulomb's Law,

$F=\dfrac{1}{4\pi \varepsilon _{0}}\dfrac{q_{1}q_{2}}{d^{2}}$

$F=\dfrac{1}{4\pi \varepsilon _{0}}\dfrac{q^{2}}{d^{2}}$

So,

$q^{2}=4\pi \varepsilon _{0}\times F\times d^{2}$

$=\dfrac{1}{9\times 10^{9}}\times 3.7\times 10^{-9}\times (5\times 10^{-10})^{2}$

$=10.28\times 10^{-38}$

$q=3.2\times 10^{-19}\, C$

As,

$q=ne$

So,

$n=\dfrac qe=\dfrac{3.2\times 10^{-19}}{1.6\times 10^{-19}}$

$n=2$

A free pith-ball of $8\ g$ carries a positive charge of $5\times 10^{-8}C$ . What must be the nature and magnitude of charge that should be given to a second pith-ball fixed $5\ cm$ vertically below the former pith-ball so that the upper pith-ball is stationary?

To keep the first pith ball stationary, the net force acting on it should cancel out to be zero.

The force of gravity acting downward

$=Mg=\dfrac{8\times10}{1000}=0.08\: N$

So, this should be the force acting on it along the upward direction due to other charged pith ball.

Hence, the pith ball placed vertically downward should exert a repulsive force on the first pith ball.

$F=0.08\: N=\dfrac{9\times10^{9}\times 5\times 10^{-8}\times q}{(0.05)^{2}}$

Hence,

$q=4.4\times 10^{-7}\: C$

Two pith-balls of same weight are suspended from the same point by means of silk threads of equal length. On charging the pith-balls equally, they are found to repel each other to a certain distance and remain at equilibrium. Identify the forces responsible for them to remain under equilibrium.

Given that two pith balls of same weight are suspended from same point by means of silk threads of equal length. On charging the pith balls equally, they repel each other. We have to find forces responsible to remain under equilibrium.

According to the figure,

At equilibrium, force of repulsion is balanced by the

$\sin$ component of tension in the string i.e,

$F=T\sin{\theta}$ ...(i)

Weight is balanced by the

$\cos$ component of the tension

$mg=T\cos{\theta}$ ...(ii)

where

$m=$ mass of each pith ball

$T=$ tension in the string

$F=$ force of repulsion

$g=$ acceleration due to gravity

Dividing (i) and (ii),

$\tan{\theta}=\dfrac{F}{mg}$

$\Rightarrow \: F=mg\tan{\theta}$

This is the force under equilibrium.

A spherical ball of mass $m$ with charge q can revolve in a vertical plane at the end of string of length $l.$ At the centre of revolution, there is a second ball with a charge identical in sign & magnitude to that of the revolving ball. What minimum horizontal velocity must be imparted to the ball in the lowest position to enable it to make a full revolution?

For the lowest most position,

$v=2\sqrt{gR}$

given,

$\therefore v=2\sqrt{gl}$ m/s

Find the electric force between two protons separated by a distance of $1$ fermi ( $1$ fermi $= {10^{ - 15}}m$ ). The protons in a nucleus remain at a separation of this order.

The distance between the two protons is

$10^{-15}\ m$ .

The amount of charge on a proton is

$1.6\times 10^{-19}\ C$

Electric force is given by

$=F_{ e }=\cfrac { k\cdot e\cdot e }{ r^{ 2 } } =\cfrac { k\cdot e^{ 2 } }{ (10^{ -15 })^{ 2 } }$

$=\cfrac { 9\times 10^{ 9 }\times (1.6\times 10^{ -19 })^{ 2 } }{ (10^{ -15 })^{ 2 } }$

$=\cfrac { 9\times 10^{ 9 }\times (1.6)^{ 2 }\times 10^{ -38 } }{ 10^{ -30 } } =90\times (1.6)^{ 2 }=230.4N$

Calculate the force between two charges each of $20 milli$ coulomb separated by a distance of $20 cm.$ in air.

Force =

$\dfrac{k q_1 q_2}{r^2}$

$= \dfrac{9 \times 10^9 \times 20 \times 10^{-3} \times 20 \times 10^{-3}}{(20 \times 10^{-2})^2}$

$= \dfrac{36 \times 10^5}{4 \times 10^{-2}}$

$= 9 \times 10^7 \, N$

If the distance between two point charges is increased by $3%$ , then calculate the $%$ decrease in the force between them.

Initial case when distance between charges is

$r$ the force can be determined as,

${ \vec { F } }_{ 1 }=\dfrac { K{ q }_{ 1 }{ q }_{ 2 } }{ { r }^{ 2 } }$

But when the radius between the charges increases that is when

$r\rightarrow 3r$ , New distance b/w charges,

${ r }^{ 1 }=3r$ , then force can be defined as,

${ \vec { F } }^{ 1 }=\dfrac { K{ q }_{ 1 }{ q }_{ 2 } }{ { \left( { r }^{ 1 } \right) }^{ 2 } } =\dfrac { K{ q }_{ 1 }{ q }_{ 2 } }{ { \left( 3r \right) }^{ 2 } } =\dfrac { K{ q }_{ 1 }{ q }_{ 2 } }{ { 9r }^{ 2 } }$

$\boxed { { \vec { F } }^{ 1 }=\dfrac { \vec { F } }{ 9 } }$

As the distance increases by 3 times the force reduces by 9 times.

Two point charges 3C and -6C are 3 m apart. Find the force between them. State the nature of the force.

${ Q }_{ 1 }=3C$

${ Q }_{ 2 }=-6C$

$r=3cm$

$K=9\times { 10 }^{ 9 }{ Nm }^{ 2 }{ C }^{ -2 }$

$F=\dfrac { K{ Q }_{ 1 }{ Q }_{ 2 } }{ { r }^{ 2 } } =\dfrac { 9\times { 10 }^{ 9 }\times 3\times \left( -6 \right) }{ 9 } =-18\times { 10 }^{ 9 }N$

$\therefore$ Altractive force due to (-ve) sin or opposite charges

A charge of $10\mu c$ and $- 10\mu c$ are placed 1cm apart. A third charge of is to be placed on the line passing through the two charges such that it is at equilibrium. Find the position of the third point. How does your answer change if the magnitude of the third charge is double?

Let the third charge be

$+Q$

Let the charges be in the equilibrium

Then the net force on

$+Q$ will be zero

$\cfrac { K\times 10\times { 10 }^{ -6 }\times Q }{ { x }^{ 2 } } =\cfrac { K\times 10\times { 10 }^{ -6 }\times Q }{ { \left( 0.01+x \right) }^{ 2 } }$

$+{ \left( 0.01+x \right) }^{ 2 }={ x }^{ 2 }$

${ 10 }^{ -4 }+{ x }^{ 2 }+0.02x=+{ x }^{ 2 }$

${ 10 }^{ -4 }=+0.02x$

$x=\cfrac { { 10 }^{ -4 } }{ 2\times { 10 }^{ -2 } }$

$x=0.5\times { 10 }^{ -2 }$

$x=0.5cm$

959419_1018637_ans_74ae5f5957b94209b8f2865e267737b6.jpg

Calculate the intensity of the electric field due to a helium nucleus at a distance of $1\overset {\circ}{A}$ from the nucleus.

2150183_1021155_ans_c6726d71fa924806855683cf80d381ca.jpeg

The force of repulsion between two point charges is $F$ , when these are at a distance of $1\ m$ apart. Now they are replaced by conducting spheres each of radii $25\ cm$ having the same charge as that of point charges. The distance between their centres is $1\ m$ , then compare the force of repulsion in the two cases.

For point charges

$F=\dfrac{kq_1q_2}{r^2}=kq_1q_2$ , since

$r=1 m$

Since charge on conducting sphere is assumed to be placed at the center of the sphere, hence repulsive force on both sphere is also given by,

$F=kq_1q_2$ , hence force on both sphere will be same as the repulsive force on the both point charges. It does not depend on the radius of the sphere.

Two point charges separated by some distance, repel each other with a force $F$ , what will be the force if the distance between them is halved?

$Force \propto \dfrac{1}{r^2}$

if distance is get halved (means 1/2) then force will get

$2^2$ times i.e

$4$ times i.e.,

$4F$

Two particles with charges $q_1$ and $q_2$ are kept at a certain distance to exert force F on the other. If the distance is reduced to one-fifth, then the force between them is :

Let the distance between them be

$d$

Force

$F = \dfrac{kq_1q_2}{d^2}$

$\implies \ F\propto\dfrac{1}{d^2}$

Si, if the distance between tem is reduced by a factor of

$5$ , force between them gets increased by a factor of

$25$ .

So, new force

$F' = 25F$

under mutual gravitational and electric force then calculate the order of $\frac{Q}{M}$ in SI units.
The force between two point charges is 100 N in air. Calculate the force if the distance between them is increased by 50%.

Since

$F_{electrostatic} = F_{gravity}$

$\dfrac{KQ^2}{r^2} = \dfrac{GM^2}{r^2}$

We get

$\dfrac{Q}{M} = \sqrt{\dfrac{G}{K}} = \sqrt{\dfrac{6.67\times 10^{-11}}{9\times 10^9}} = 8.6\times 10^{-11}$

Let the initial distance between the charges be

$d$ .

New distance between them

$d_2 = 1.5 \ d$

Force

$F = \dfrac{KQ^2}{d^2}$

$\implies$

$\dfrac{F_2}{F_1} =( \dfrac{d_1}{d_2})^2$

$\dfrac{F_2}{100} =( \dfrac{d}{1.5d})^2 = \dfrac{1}{2.25}$

$\implies \ F_2 = 44.44 \ N$

Two charge are placed at some separation in vacuum experiences a force $F$ . If a charge are halfed with separation double and are placed in a medium of dielective constant $4$ . Find the force experience b the

$F = \dfrac{1}{{4\pi \in o}} \times \dfrac{{{q_1}{q_2}}}{{{r^2}}}$

If the charges are halfed and the seperation is doubled

then

$F' = 4 \times \frac{1}{{4\pi \in o}} \times \dfrac{{\dfrac{{{q_1}}}{2} \times \dfrac{{{q_2}}}{2}}}{{{{\left( {2r} \right)}^2}}} = \dfrac{1}{4} \times \dfrac{1}{{4\pi \in o}} \times \dfrac{{{q_1}{q_2}}}{{{r^2}}}$

$= \dfrac{1}{4}F$

A charge q=1 $\mu$ C is placed at point (1m,2m,4m),. find the electric field at point P(0m,-4m,3m).

$\textbf{Step 1: Calculation of r [Ref. Fig.]}$

Let

$r$ be the distance between point

$(1,2,4)m$ and

$P(0,-4,3)m$

$\vec{r}=\vec{r_P}-\vec{r_Q}=(0\hat{i}-4\hat{j}+3\hat{k})-(1\hat{i}+2\hat{j}+4\hat{k})$

$\Rightarrow \vec{r}=-1\hat{i}-6\hat{j}-\hat{k}$

Magnitude of

$r:$

$r = \sqrt{(0-1)^2 + (-4-2)^2 + (3-4)^2}$

$\Rightarrow r = \sqrt{1 + 36 + 1} = \sqrt{38} = 6.16 m$

$\textbf{Step 2: Electric field at point P }$

Electric field due to point charge can be given as:

$\vec{E} = \dfrac{1 }{4 \pi \epsilon_0} \dfrac{q\vec{r}}{r^3} = \dfrac{9 \times 10^9 \times 1 \times 10^{-6}C}{(6.16\ m)^3}$

$(-1\hat{i}-6\hat{j}-\hat{k})$

$\Rightarrow \vec{E} = 38.5$

$(-1\hat{i}-6\hat{j}-\hat{k})N/C$

$|\vec{E}|=\sqrt{(38.5)^2[(-1)^2+(-6)^2+(-1)^2]} \ N/C$

$=237.3 N/C$

2116653_1215238_ans_a8a50d2b72314d8281b7e6b88abe7e6a.png

Two charges of value $2 \ \mu C$ and $-50 \mu C$ are placed 80 cm apart. Calculate the distance of the point from the smaller charge where the intensity is zero.

Amount of first charge is $=2\times { 10 }^{ -6 }C$

Amount of second charge is $=50\times { 10 }^{ -6 }C$

Distance $= 80 cm$

Electric Field $=\frac { k{ q }^{ 1 }{ q }^{ 2 } }{ r }$

Electric Field will be zero

E.F. $=\dfrac { 0(K{ x }^{ 2 }\times 10^{ - }6\times 50\times 10^{ - }6) }{ 80 } +\left( \dfrac { k\times 2\times { 10 }^{ -6 } }{ r } \right)$

$=0\left( \dfrac { { 10 }^{ -4 } }{ 80 } \right) +\left( 2\times \dfrac { { 10 }^{ -6 } }{ r } \right)$

$=0\left( \dfrac { { 10 }^{ -4 } }{ 80 } \right)$

$=-\left( 2\times \dfrac { { 10 }^{ -6 } }{ r } \right) \left( \dfrac { { 10 }^{ 2 } }{ 80 } \right)$

$=\dfrac { 2 }{ r\left( \dfrac { 10 }{ 8 } \right) }$

$=\dfrac { 2 }{ r\left( \dfrac { 10 }{ 4 } \right) }$

$=\dfrac { 1 }{ r }$

$R=0.4$

Two equal charges are placed in air are separated by a distance of 3 meter ,repel each other with a force of $0.19\,f$ .calculate the magnitude of either of the charge ( $9.9 \times 10^{-1}C)$

Given,

$d=3m,f=0.19f$

So,

$f=\dfrac{kq^2}{r^2}\Rightarrow kq^2=0.19\times9\Rightarrow q^2=\dfrac{1.71}{9.0\times10^9}$ Where

$k=9.0\times 10^9$

$\Rightarrow q=\sqrt{0.19\times10^{-9}}$ C

The point-like charges carrying charges of $+3\times 10^{-9} C$ and $-5\times 10^{-9} C$ are $2 \ m$ apart. Determine the magnitude of the force between them and state whether it is attractive or repulsive.

We use coulomb's law to calculate the magnitude of the force.

$F=k\dfrac{q_{1}q_{2}}{r^2}$

$q_{1}=+3\times 10^{-9} \ C$

$q_{2}=-5\times 10^{-9} \ C$

$r = 2 \ m$

$k=9\times 10^9 \ N \ m^2 \ C^{-2}$

Therefore,

$F=\dfrac{9\times 10^9 \times 3 \times 10^{-9} \times 5 \times 10^{-9}}{4}$

$F=3.37 \times 10^{-8} \ N$

Since, the point charges have opposite signs, the force will be attractive.

The force between two electrons when placed in air is equal to $0.5$ times the weight of an electron. Find the distance between two electrons (mass of electron $=9.1\times {10}^{-31}kg$ )

Electrostatic force

$F_e = \dfrac{ke^2}{r^2}$

Weight of electron

$W = mg$

So,

$F_e = 0.5 W$

$\dfrac{ke^2}{r^2} = 0.5 mg$

$\dfrac{9\times 10^9(1.6\times 10^{-19})^2}{r^2} = 0.5\times 9.1\times 10^{-31}\times 10$

$\implies \ r = 7.2 \ m$

A charge $Q$ is to be divided on two objects.What should be the
values of the charges on the two objects so that the force between the objects
can be maximum?

1328417_1113108_ans_9f15f1cc509d463693cf89218a044d86.jpg

Two identical small conducting spheres carry charges of $Q_1$ and $Q_2$ with $Q_1 >> Q_2$ . The spheres are d distance apart. The force they exert on each other is $F_1$ . The spheres are made to touch each other and then separated by a distance d. The force they exert on each other now is $F_2$ . Then $F_1/F_2$ is :-

In the

$1$ st case we have:

$F_1=\dfrac{kQ_1Q_2}{d^2}$

$($ By coulomb's law,

$k=\dfrac{1}{4\pi\epsilon_0}=9\times 10^9Nm^2/C^2)$

After touching the spheres the charges distribute such that charge on sphere is:

$q=\dfrac{Q_1+Q_2}{2}\approx\dfrac{Q_1}{2}(Q_1 >> Q_2)$

So new force between them will be:

$F_2=\dfrac{kq^2}{d^2}=\dfrac{kQ^2_1}{4d^2}$

Hence

$\dfrac{F_1}{F_2}=\dfrac{4Q_1Q_2}{Q^2_1}=\dfrac{4Q_2}{Q_1}$ .

Two point charges, five time as strong as the other repel each other with a force of $1.8 \times 10^3$ N. When separated by a distance 1 m, calculate the value of charges?

it is given that,

${{q}_{1}}=5{{q}_{2}}.............(1)$

$F=1.8\times {{10}^{3}}\,N$

$r=1\,\,m$

According to coulomb’s law

$F=k\dfrac{{{q}_{1}}{{q}_{2}}}{{{r}^{2}}}$

$1.8\times {{10}^{3}}=9\times {{10}^{9}}\times \dfrac{5{{q}_{2}}{{q}_{2}}}{1}\,\,\,\,\,(from\,\,(1)\,)$

$1.8\times {{10}^{3}}=45\times {{10}^{9}}\times {{q}_{2}}{{q}_{2}}$

$q_{2}^{2}=4\times {{10}^{-8}}$

${{q}_{2}}=2\times {{10}^{-4}}\,C$

Put this value in equation (1)

${{q}_{1}}={{10}^{-3}}\,C$

Three equal charges, $2\cdot 0\times { 10 }^{ -6 }C$ each, are held fixed at the three corners of an equilateral triangle, of side $5\ cm$ . Find the Coulomb force experienced by one of the charges, due to the other two.

1309868_1138246_ans_31f87e9b594b48909bd8e4801ed2adf0.jpg

A metal plate of area $0.01\ m^{2}$ carries a charge of $100\ \mu C$ . Calculate the outward pull on one side of the plate. $[k=1]$

1309693_1137732_ans_ee0152682fdc4a2f85aedf599a9231bd.jpg

Two point charges of $+16\mu C$ and $-9\mu C$ are placed $8\ cm$ apart in air. Determine the position of the point at which the resultant field is zero.

$E_{1} = \dfrac {kq_{1}}{x^{2}} = \dfrac {k\times q}{x^{2}}$

$E_{2} = \dfrac {kq_{2}}{(8 + x)^{2}} = \dfrac {k\times 16}{(x + 8)^{2}}$

$E_{1} = E_{2}$

$\dfrac {9k}{x^{2}} = \dfrac {16k}{(x + 8)^{2}}$

$\dfrac {3}{x} = \dfrac {4}{x + 8}$

$4x = 3x + 24$

$x = 24$ .
2080083_1237213_ans_899ee213c5414e09ab8b51462cb2b775.jpg

Plot a graph showing the variation of coulomb force $(F)$ versus $\left (\dfrac {1}{r^{2}}\right )$ where $r$ is the distance between the two charges of each pair of charges : $(1\mu C, 2\mu C)$ and $(2\mu C, -3mu C)$ . Interpret the graphs obtained.

$F = \dfrac {1}{4\pi \epsilon_{0}} \dfrac {q_{1}q_{2}}{r^{2}}$
The graph between

$F$ and

$\dfrac {1}{r^{2}}$ is a straight line of slope

$\dfrac {1}{4\pi \epsilon_{0}} q_{1}q_{2}$ passing through origin in both the cases.
Since, magnitude of the slope is more for attention, therefore, attractive force is greater than repulsive force.
1663100_1230013_ans_c96dcf59a76b46609e4abb6bdfd5a151.png

1663100_1230013_ans_c96dcf59a76b46609e4abb6bdfd5a151.png

A particle of charge $2 \mu C$ and mass $1.6 g$ is moving with a velocity $4 \hat{i} ms^{-1}$ . At $t = 0$ the particle enters in a region having an electric field $\vec{E}$ (in $NC^{-1}) = 80 \hat{i} + 60 \hat{j}$ . Find the velocity of the particle at $t = 5 s$ .

Given that :-

q = charge

$= 2 \mu C = 2 \times 10^{-6} C$

mass

$= m = 1.6 g = 1.6 \times 10^{-3} kg$

$V =$ velocity

$= 4 \hat{i} m/s$

$\vec{E} = (80 \hat{i} + 60 j ) N/C$

Since particle has initial velocity only in x-direction

$u_x = v = 4 m/s ; \, u_y = 0 m/s$

Force

$= F = q \vec{E}$

$\Rightarrow \vec{F} = 2 \times 10^{-6} (80 \hat{i} + 60 j) = (160 \hat{i} + 120 j) \times 10^{-6}$

$\vec{a} =$ acceleration

$= \vec{a} = \dfrac{\vec{F}}{m} = \dfrac{(160 \hat{i} + 120 j) \times 10^{-6}}{1.6 \times 10^{-3}}$

$\vec{a} = (100 \hat{i} + 75 \hat{j}) \times 10^{-3} = (a_x \hat{i} + a_y \hat{j})$

time

$= t = 5 sec$

$v_x = u_x + a_x t = (4 + 100 \times 10^{-3} \times 5)$

$v_x = 4 + 0.5 = 4.5 m/s$

$v_y = u_y + a_y t = 0 + 75 \times 10^{-3} \times 5 = 0.375 m/s$

Where

$u_x , v_x$ are initial and final velocity in x-direction and

$u_y$ and

$v_y$ are initial and final velocity in y-direction.

$\therefore$ Final velocity

$= v = v_x \hat{i} + v_y \hat{j} = 4.5 \hat{i} + 0.375 j$

$v = (4.5 \hat{i} + 0.375 \hat{j}) m/s$

1497126_1700248_ans_64d273e977d8467cad52d682e344afb1.png

Two point charges of $+1 \mu C$ and $+4 \mu C$ are kept $30 cm$ apart. How far from the $+1 \mu C$ charge on the line joining the two charge, will the net electric field be zero ?

Given ,

Two charge

$q_1 = +1\mu C$

$q_p = 1\mu C$

$q_2 = 4\mu C$

$d = 30cm = .3m$

Let point

$P$ (where,

$f_{1\mu C}=0$ ) is at distance

$r$ from

$q_1$ [ref. image]

$F_1+ F_2 = 0$

$k \dfrac{q_1 \times q_p}{r^2} - k \dfrac{q_1q_p}{(.3-r)^2} = 0$

$\dfrac{q_1}{r^2} = \dfrac{q_2}{(*.-2)^2}$

$\left(\dfrac{.3-r}{r}\right)^2 = \dfrac{q_2}{q_1} = \dfrac{4}{1} = 4$

$\left( \dfrac{.3-r}{r}\right)^2 = 4$ or

$\dfrac{.3-r}{r} = 2$

$\dfrac{.3-r}{r} = +2$

$r = .1m = 10cm$

1497056_1700272_ans_d5552bc57e144ed8aa0fac5444cde3e2.png

A pendulum bob of mass $80$ mg and carrying a charge of $2\times 10^{-8}$ C is at rest in a uniform, horizontal electric field of $20 kVm^{-1}$ . Find the tension in the thread.

The electric field in the region is

$E=20 kv/m=20\times 10^3V/m$ .

The mass of the pendulum bob is

$m=80\times 10^{-5}kg$

The charge on the pendulum bob is

$q=2\times 10^{-8}$ C

At equilibrium, the force of the electric charge on the bob will be balanced by the tension and the weight of the bob.

The force balancing equations for the bob are as:

$T\sin\theta =qE$

$T\cos \theta =mg$

The value of the angle made by the tension is:

$\tan\theta =\left(\dfrac{qE}{mg}\right)$

$\tan\theta =\left(\dfrac{2\times 10^{-8}\times 20\times 10^3}{80\times 10^{-6}\times 10}\right)=\left(\dfrac{1}{2}\right)$

$\theta\approx 27^\circ$

Now, tension can be obtained as:

$T=\dfrac{qE}{cos\theta}$

$=\dfrac{2\times 10^{-8}\times20\times 10^3 }{cos 27^\circ}$

$8.8\times 10^{-4}\ N$

1563665_1720238_ans_43b232f939ba497a801dc4ee31ad8b99.png

A particle of mass m and charge q is thrown at a speed u against a uniform electric field E. How much distance will it travel before coming to momentary rest?

Given

$u=$ Velocity of projection,

$\vec{E}=$ Electric field intensity

$q=$ charge on the particle

$m=$ mass of particle

We know, Force experienced by a particle with charge 'q' in an electric field

$\vec{E}=qE$

$\therefore$ the acceleration produced is given as:

$a=\dfrac{qE}{m}$ .

As the particle is projected against the electric field, hence deceleration

$=\dfrac{qE}{m}$
So, let the distance covered be 's'

Then,

$v^2=u^2+2as$

[where

$a=$ acceleration,

$v=$ final velocity]

$0=u^2-2\times \dfrac{qE}{m}\times S$

$\Rightarrow S=\dfrac{u^2m}{2qE}$ units.

A particle having a charge of $2.0\times 10^{-4}$ C is placed directly below and at a separation of $10$ cm from the bob of a simple pendulum at rest. The mass of the bob is $100$ g. What charge should the bob be given so that the string becomes loose?

Mass of the bob

$=100$ g

$=0.1$ kg

So Tension in the string

$=0.1\times 9.8=0.98$ N.

For the Tension to be

$0$ , the charge below should repet the first bob.

$\Rightarrow F=\dfrac{kq_1q_2}{r^2}$

At equilibrium, the forces acting on the particle is:

$T-mg+F=0$

When spring becomes loose, the tension will be zero.

$\Rightarrow 0=mg-F$

$F=mg$

$\Rightarrow 0.98=\dfrac{9\times 10^9\times 2\times 10^{-4}\times q_2}{(0.01)^2}$

$\Rightarrow q_2=\dfrac{0.98\times 1\times 10^{-4}}{9\times 2\times 10^5}=0.054\times 10^{-9}N$ .

1562393_1720165_ans_113de0606ec2484382fc697035ae6f2c.png

A particle of mass $1$ g and charge $2.5\times 10^{-4}$ C is released from rest in an electric field of $1.2\times 10^{4}NC^{-1}$ . How long will it take for the particle to travel a distance of $40$ cm?

The mass of the particle is

$m=1g=10^{-3}kg$

The initial velocity of particle is

$u=0$

The charge on particle is

$q=2.5\times 10^{-4}C$

The magnitude of electric field is

$E=1.2\times 10^4N/C$

The distance covered by the

$S=40cm=4\times 10^{-1}m$

The acceleration of the charged particle is :

$a=\dfrac{qE}{m}$

$=\dfrac{2.5\times 10^{-4}\times1.2\times 10^4}{10^{-3}}$

$=3\times 10^3\ m/s^2$

the time taken by the charge to cover the distance is:

$S=\dfrac{1}{2}gt^2$

$t=\sqrt{\dfrac{2S}{g}}$

$=\sqrt{\dfrac{2\times 4\times 10^{-1}}{3\times 10^3}}=1.63\times 10^{-2}$ sec.

A ball of mass $100$ g and having a charge of $4.9\times 10^{-5}$ C is released from rest in a region where a horizontal electric field of $2.0\times 10^4N C^{-1}$ exists. Find the resulatnt force acting on the ball.

The mass of ball is

$m=100g$

The charge on the ball is

$q=4.9\times 10^{-5}\ C$

The magnitude of electric field intensity is

$\vec{E}=2\times 10^{4}N/C$

The gravitational forcve on the ball is

$F_g=mg$

The force on the ball due to electric field is

$F_e=qE$

So, the practicle moves due to the net resultant R

$R=\sqrt{F^2_g+F^2_e}$

$=\sqrt{(0.1\times 9.8)^2+(4.9\times 10^{-5}\times 2\times 10^{4})^2}$

$=\sqrt{0.9604+96.04\times 10^{-2}}=\sqrt{1.9208}=1.3859N$ .

1563678_1720252_ans_57b81854b5514264a1098abb8019ede4.png

A long cylindrical volume contains a uniformly distributed charge of density $\rho$ . Find the electric field at a point $P$ inside the cylindrical volume at a distance $x$ from its axis.

Given : Volume charge density

$= \zeta$

Let the height of cylinder be

$h$ .

$\therefore$ Charge

$Q$ at

$P=\zeta \times 4\pi x^2 \times h$

For electric field

$\oint E \ ds = \dfrac{Q}{\varepsilon_0}$

$E = \dfrac{Q}{\varepsilon_0 \times dx} = \dfrac{\zeta \times 4\pi x^2 \times h}{\varepsilon_0 \times 2 \times \pi \times x \times h} = \dfrac{2\zeta x}{\varepsilon_0}$

1545735_1720263_ans_64e40553604c4ac49c5871b48bf7ac52.png

A particle of mass $1$ g and charge $2.5\times 10^{-4}$ C is released from rest in an electric field of $1.2\times 10^{4}NC^{-1}$ . What will be the speed of the particle after travelling the distance of 40cm?

1604632_1720250_ans_5d56a4c4a9ef4910ae4c9dea1d21414d.jpg

Two particles of charges and masses $(+q_1, m_1)$ and $(-q_2, m_2)$ are released at different locations in a uniform electric field E in free space. If their separation remains unchanged, find the separation between them.

A copper atom consists of copper nucleus surrounded by $29$ electrons. The atomic weight of copper is $63.5g$ ${mol}^{-1}$ . Let us now take two pieces of copper each weighing $10g$ . Let us consider one electron from one piece is transferred to another from every $1000$ atoms in a piece.
(a) Find the magnitude of charge appearing on each piece
(b) What will be the coulomb force between the two pieces after the transfer of electrons if they are $10cm$ apart?
(Avagadro's number $=6\times {10}^{23}$ ${mol}^{-1}$ )

No. of atoms in

$10g$ of copper

$n=\cfrac { 10 }{ 63.5 } \times 6\times { 10 }^{ 23 }=9.448\times { 10 }^{ 22 }\quad$
Number of electrons transferred,

$N=\cfrac { n\times 1 }{ 1000 } =9.448\times { 10 }^{ 19 }$
(a) Magnitude of charge appearing on either piece

$q=Ne=9.448\times { 10 }^{ 22 }\times 1.6\times { 10 }^{ -19 }=15.12C\quad$

(b)

$F=k\cfrac { { q }^{ 2 } }{ { r }^{ 2 } } =\cfrac { 9\times { 10 }^{ 9 }\times { (15.12) }^{ 2 } }{ { \left( 0.10 \right) }^{ 2 } } =2.05\times { 10 }^{ 14 }N\quad \quad$

A ball of mass $100$ g and having a charge of $4.9\times 10^{-8}$ C is released from rest in a region where a horizontal electric field of $2.0\times 10^4N C^{-1}$ exists. Where will the ball be at the end of $2s$ ?

The mass of ball is

$m=100g$

The charge on the ball is

$q=4.9\times 10^{-5}\ C$

The magnitude of electric field intensity is

$\vec{E}=2\times 10^{4}N/C$

The gravitational forcve on the ball is

$F_g=mg$

The force on the ball due to electric field is

$F_e=qE$

So, the practicle moves due to the net resultant R

$R=\sqrt{F^2_g+F^2_e}$

$=\sqrt{(0.1\times 9.8)^2+(4.9\times 10^{-5}\times 2\times 10^{4})^2}$

$=\sqrt{0.9604+96.04\times 10^{-2}}=\sqrt{1.9208}=1.3859N$ .

The angle between

$\tan\theta =\dfrac{F_g}{F_e}=\dfrac{mg}{qE}=1$

So,

$\theta =45^o$

$\therefore$ Hence path is straight along resultant force at an angle

$45^o$ with horizontal

Disp. Vertical

$=(1/2)\times 9.8\times 2\times 2=19.6$ m

Disp. Horizontal

$=S=(1/2)at^2=\dfrac{1}{2}\times \dfrac{qE}{m}\times t^2=\dfrac{1}{2}\times \dfrac{0.98}{0.1}\times 2\times 2=19.6m$

Net Dispt.

$=\sqrt{(19.6)^2+(19.6)^2}=\sqrt{768.32}=27.7m$ .

1563877_1720254_ans_40feb28a4c9647f882297cbd80f914bf.png

A particle of mass $1$ g and charge $2.5\times 10^{-4}$ C is released from rest in an electric field of $1.2\times 10^{4}NC^{-1}$ . How much is the work done by the electric force on the particle as it moves for 40cm?

1604643_1720251_ans_95c3a3dad119401380224014b15a83ea.PNG

A grounded metallic ball of radius r is placed on the centre of a uniformly charged thin insulating disc of radius $R (R >> r).$ If total charge on the Disc is Q, find force of electrostatic interaction between the ball and the disc.

1674449_1731978_ans_4d02f08b55914422844a95389dcfc726.jpg

A ball of mass $100$ g and having a charge of $4.9\times 10^{-5}$ C is released from rest in a region where a horizontal electric field of $2.0\times 10^4N C^{-1}$ exists. What will be the path of the ball?

The mass of ball is

$m=100g$

The charge on the ball is

$q=4.9\times 10^{-5}\ C$

The magnitude of electric field intensity is

$\vec{E}=2\times 10^{4}N/C$

The gravitational forcve on the ball is

$F_g=mg$

The force on the ball due to electric field is

$F_e=qE$

So, the practicle moves due to the net resultant R

$R=\sqrt{F^2_g+F^2_e}$

$=\sqrt{(0.1\times 9.8)^2+(4.9\times 10^{-5}\times 2\times 10^{4})^2}$

$=\sqrt{0.9604+96.04\times 10^{-2}}=\sqrt{1.9208}=1.3859N$ .

Now the anhgle will be:

$\tan\theta =\dfrac{F_g}{F_e}=\dfrac{mg}{qE}=1$

So,

$\theta =45^o$

$\therefore$ Hence the path is straight along with resultant force at an angle

$45^o$ with horizontal.

1563681_1720253_ans_14dd184fc8fe4ea1977fe3720cf2ecea.png

Three particles, each with positive charge $Q$ , form an equilateral triangle, with each side of length $d$ . What is the magnitude of the electric field produced by the particles at the midpoint of any side?

The two closest charges produce fields at the midpoint that cancel each other out.

Thus, the only significant contribution is from the furthest charge,

which is a distance

$r=\sqrt 3 d/2$ away from that midpoint.

Plugging this immediately gives the result:

$E=\dfrac{Q}{4\pi \varepsilon_0r^2}$

$=\dfrac{Q}{4\pi \varepsilon_0 (\sqrt 3d/2)^2}$

$=\dfrac{4}{3}\dfrac{Q}{4\pi \varepsilon_0 d^2}$ .

What is the acceleration of an electron in an uniform electric field of magnitude $1.40\times 10^6\ N/C$ ?

We know,

$\Delta x=\dfrac{v^2-v_0^2}{2a}$

$=\dfrac{(3.00\times 10^7\ m/s)^2}{2(2.46\times 10^{17}\ m/s^2)}$

$=1.83\times 10^{-3}\ m$ .

Two tiny spherical water drops with identical charges of $-1.00\times 10^{-16}C$ , have a center-to-center separation of $1.00\ cm$ .
What is the magnitude of the electrostatic force acting between them?

Equation

$21-1$ gives

$F=\dfrac{Kq_1 q_2}{r^2}$

$F=\dfrac{(8.99\times 10^{9}N.m^{2}/C^{2})(1.00\times 10^{-16}C)^{2}}{(1.00\times 10^{-2}m)^{2}}=8.99\times 10^{-19}N$

What is the magnitude of the electrostatic force between a singly charged sodium ion ( $Na^{+}$ , of charge $+e$ ) and an adjacent singly charged chlorine ion ( $Cl^{-}$ , of charge $-e$ ) in a salt crystal if their separation is $2.82\times 10^{-10}m$ ?

The magnitude of the force is

$F=k\dfrac{e^{2}}{r^{2}}=\left(8.99\times 10^{9}\dfrac{N.m^{2}}{C^{2}}\right)\dfrac{(1.60\times 10^{-19}C)^{2}}{(2.82\times 10^{-10}m)^{2}}=2.89\times 10^{-9}N$

A charge of $6.0 \mu C$ is to be split into two parts that are then separated by $3.0\ mm.$ What is the maximum possible magnitude of the electrostatic force between those two parts?

Let

$q_1$ be the charge of one part and

$q_2$ that of the other part ; thus,

$q_1+q_2=Q=6.0\ \mu C$ .The repulsive force between them is given by Coulomb's law:

$F=\dfrac{q_1q_2}{4\ \pi \varepsilon _0 r^2} =\dfrac{q_1(Q-q_2)}{4\ \pi \varepsilon _0 r^2}$
If we maximize this expression by taking the derivative with respect to

$q_1$ and setting equal to zero, we find

$1_1=Q/2$ , which might have been anticipated (based on symmetry arguments). This implies

$q_2=Q/2$ also. With

$r=0.0030\ m$ and

$Q=6.0 \times 10^{-6}\ C$ , we find

$F=\dfrac{(Q/2)(Q/2)}{4\ \pi \varepsilon _0 r^2}=\dfrac{1}{4}\dfrac{Q}{4\ \pi \varepsilon _0 r^2}=\dfrac{1}{4}\dfrac{(8.99 \times 10^9\ N/m^2/C^2)(6.0 \times 10^{-6}\ C)^2}{(3.00 \times 10^{-3}m)^2} \approx 9.0 \times 10^3\ N.$

The charge and coordinates of two charged particles held fixed in an $y$ plane are $q_1=+3.0\ \mu C, x_1=3.5\ cm, y_1=0.50\ cm$ , and $q_2 =-4.0\mu C, x_2 =-2.0\ cm, y_2 =1.5\ cm$ . Find
magnitude?

The distance between

$q_1$ and

$q_2$ is

$r_{12}=\sqrt {(x_2 -x_1)^2 +(y_2 -y_1)^2}=\sqrt {(-0.020\ m-0.035\ m)^2 +(0.015\ m-0.005\ m)^2}=0.56\ m$
The magnitude of the force exerted by

$q_1$ on

$q_2$ is

$F_{21}=k\dfrac {|q_1 q_2|}{r_{12}^2}=\dfrac {(8.99\times 10^9 N.m^2 /C^2)(3.0\times 10^{-6}C)(4.0\times 10^{-6}C)}{(0.56\ m)^2}=35\ N$ .

What would be the magnitude of the electrostatic force between two $1.00\ C$ point charges separated by a distance of $1.00\ km$ if such point charges existed (they do not) and this configuration could be set up?

$r=1000\ m$ , then

$F=\dfrac{q_1q_2}{4\ \pi \varepsilon _0 r^2}=\dfrac{kq^2}{r^2}=\dfrac{(8.99 \times 10^9\ N.m^2/C^2)(1.00\ C)^2}{(1.00 \times 10^3\ m)^2}=8.99 \times 10^3\ N$ .

In figure, particle $1$ and $2$ have charge $20.0\ \mu C$ each and are held at separation distance $d=1.50\ m$ .
What is the magnitude of the electrostatic force on particle $1$ due to particle $2$ ? In figure, particle $3$ of charge $20.0\ \mu C$ is positioned so as to complete an equilateral triangle?

Equation gives

$F_{12}=k\dfrac {q_1 q_2}{d^2}=(8.99\times 10^9 N.m^2 /C^2)\dfrac {(20.0\times 10^{-6}C)^2}{(1.05m)^2}=1.60\ N$

What would be the magnitude of the electrostatic force between two $1.00\ C$ point charges separated by a distance of $1.00\ m$

Using Coulomb's law, we obtain

$F=\dfrac{q_1q_2}{4\ \pi \varepsilon _0 r^2}=\dfrac{kq^2}{r^2}=\dfrac{(8.99 \times 10^9\ N.m^2/C^2)(1.00\ C)^2}{(1.00\ m)^2}=8.99 \times 10^9\ N$ .

A particle of charge $+3.00\times 10^{-6}C$ is $12.0\ cm$ distant from a second particle of charge $-1.50\times 10^{-6}C$ . Calculate the magnitude of the electrostatic force between the particles.

The magnitude of the mutual force of the attraction at

$r=0.120\ m$ is

$F=k\dfrac {|q_1||q_2|}{r^2}(8.99\times 10^9 N-m^2/C^2)\dfrac {(3.00\times 10^6 C)(1.50\times 10^{-6}C)}{(0.120m)^2}=2.81\ N$

In figure, eight particles form a square in which distance $d=2.0\ cm$ . The charges are $q_1=+3e, q_2=+e, q_3=-5e, q_4=-2e, q_5=+3e, q_6=+w, q_7=-5e$ , and $q_8=+e$ . In univector notation, what is the net electric field at the square's centre?

Most of the individual fields, caused by diametrically opposite charges, will cancel, except for the pair that lie on the

$x$ axis passing through the center. The pair of charges produces a field pointing to the right

$\vec E$

$=\dfrac{3q}{4\pi \varepsilon_0 d^2}\hat i$

$=\dfrac{3e}{4\pi \varepsilon_0 d^2}=\hat i$

$=\dfrac{3(8.99\times 10^9\ N.m^2/C^2)(1.60\times 10^{-19}\ C)}{(0.020\ m)^2}\hat i =(1.08\times 10^{-5}\ N/C) \hat i$ .

In Fig. Three identical conducting spheres form an equilateral triangle of side length $d=20.0\ cm$ . The sphere radii are much smaller than $d,$ and the sphere charges are $q_A=-2.00\ nC, \ q_B=-4.00\ nC,\ and\ a_C=+8.00\ nC$ . between spheres B and C?

We also obtain

$|\xrightarrow { F } _{AC}|=\dfrac{|q_A q_C|}{4 \pi \varepsilon _0 d^2}=\dfrac{|(8.99 \times 10^9\ N.m^2/C^2)(-4.00 \times 10^{-9}\ C)(-4.00 \times 10^{-9}\ C)|}{(0.200\ m)^2}=2.70 \times 10^{-6}\ N$ .

A neutron consists of ine "up" quark of charge $2e/3$ and two "down" quarks each having charge $-e/3.$ If we assume that the down quarks are $26 \times 10^{-15}$ \ m$$ apart inside the neutron, what is the magnitude of the electrostatic force between them?

Coulomb's law gives

${ F } =\dfrac{|q|^2}{4 \pi \varepsilon _0 d^2}=\dfrac{k(e/3)^2}{r^2}=\dfrac{(8.99 \times 10^9\ N.m^2/C^2)(1.60 \times 10^{-19}\ C)}{9(2.6 \times 10^{-15\ m})^2}=3.8\ N$ .

Figure shows a circular disk that is uniformly charged. The central $z$ axis is perpendicular to the disk face, with the origin at the disk. Figure gives the magnitude of the electric field along that axis in terms of the maximum magnitude $E-m$ at the disk surface. The $z$ axis scale is set by $z_s=8.0\ cm$ . What is the radius of the disk?

We know,

$\dfrac{E}{E_{max}}=1-\dfrac{z}{(z^2+R^2)^{1/2}}$
and note that this ratio is

$\dfrac{1}{2}$

( according to the graph shown in the figure)

when

$z=4.0\ cm$ .
Solving this for

$R$ we obtain

$R=z\sqrt 3=6.9\ cm$ .

In Fig, the three particles are fixed in place and have charges $q_{1}=q_{2}=+e$ and $q_{3}=+2e$ . Distance $a=6.00 \mu m$ . What are the
magnitude

By symmetry we see that the contributions from the two charges

$q_{1}=q_{2}=+e$ cancel each other,

and we know that to compute magnitude of the field due to

$q_{3}=+2e$
The magnitude of the net electric field is

$|\vec{E_{net}}|=\dfrac{1}{4\pi\varepsilon_{0}}\dfrac{2e}{r^{2}}=\dfrac{1}{4\pi\varepsilon_{0}}\dfrac{2e}{(a/\sqrt{2})^{2}}=\dfrac{1}{4\pi\varepsilon_{0}}\dfrac{4e}{a^{2}}$

$=(8.99\times 10^{9}N.m^{2}/C^{2})\dfrac{4(1.60\times 10^{-19}C)}{(6.00\times 10^{-6}m)^{2}}$

$=160\ N/C$

Of the charge $Q$ on a tiny sphere, a fraction $\alpha$ is to be transferred to a second, nearby sphere. The spheres can be treated as particle. What value of $\alpha$ maximizes the magnitude $F$ of the electrostatic force between the two spheres? What are the

The two charges are

$q= \alpha Q$ (where

$\alpha$ is a pure number presumably less than

$1$ and greater than zero) and

$Q-q=(1- \alpha)Q$ . Thus,Eq. gives

$F=\dfrac{1}{4\ \pi \varepsilon _0}\dfrac{(\alpha Q)((1- \alpha)Q)}{d^2}=\dfrac{Q^2 \alpha(1- \alpha)}{4\ \pi \varepsilon _0 d^2}$ .
The graph below, of

$F$ versus

$\alpha$ , has been scaled so that the maximum is

$1$ . In actuality,the maximum value of the force is

$F_{max}=Q^2/16 \pi \varepsilon _0 d^2$ .
It is clear that

$\alpha =1/2=0.5$ gives the maximum values of

$F$ .

1697066_1770956_ans_9b76b764191c46269c97f8ba70ab37d0.png

Two charged particles are attached to an $x$ axis: Particle $1$ of charge $-2.00\times 10^{-7}C$ is at position $x=6.00\ cm$ and particle $2$ of charge $+2.00\times 10^{-7}C$ is at position $x=21.0\ cm$ . Midway between the particle, what is their net electric field in unit vector notation?

With

$x_{1}=6.00\ cm$ and

$x_{2}=21.00\ cm$ ,

the point midway between the two charges is located at

$x=13.5\ cm$ .

The values of the charge are

$q_{1}=-q_{2}=-2.00\times 10^{-7}C$ .
and the magnitude and direction of the individual fields are given by:

$\vec{E_{1}}=-\dfrac{|q_{1}|}{4\pi\varepsilon_{0}(x-x_{1})^{2}}\hat{i}=-\dfrac{(8.99\times 10^{9}N.m^{2}/C^{2})|-2.00\times 10^{-7}C|}{(0.135\ m-0.060\ m)^{2}}\hat{i}$

$=-(3.196\times 10^{5}\ N/C)\hat{i}$

$\vec{E_{2}}=-\dfrac{q_{2}}{4\pi\varepsilon_{0}(x-x_{1})^{2}}\hat{i}=-\dfrac{(8.99\times 10^{9}N.m^{2}/C^{2})(-2.00\times 10^{-7}C)}{(0.135\ m-0.210\ m)^{2}}\hat{i}$

$=-(3.196\times 10^{5}\ N/C)\hat{i}$
Thus the net electric field is

$\vec{E}_{net}=\vec{E_{1}}+\vec{E_{2}}=-(6.39\times 10^{5}N/C)\hat{i}$

A charged cloud system produces an electric field in the air near Earth's surface. A particle of charge $-2.0\times 10^{-9}\ C$ is acted on by a downward electrostatic force of $3.0\times 10^{-6}\ N$ when placed in this field.
What is the magnitude of the gravitational force $\vec F_g$ on the proton?

The magnitude of the gravitational force on the proton is

$F_g$

$=mg=(1.67\times 10^{-27}\ kg)(9.8\ m/s^2)$

$=1.6\times 10^{-26}\ N$ .

The force between two point charges kept at a distance $r$ apart in air is $F$ . It the same charges are kept in water at the same distance, how does the force between them change?

As we know,

The force is air

$F_{a} = \dfrac {1}{4\pi \epsilon_{0}}\dfrac {q_{1}q_{2}}{r^{2}}$

The force in water

$F_{w} = \dfrac {1}{4\pi \epsilon_{0}K} \dfrac {q_{1}q_{2}}{r^{2}}$

$\therefore \dfrac {F_{w}}{F_{a}} = \dfrac {1}{K}$
Dielectric constant of water is

$81$ , so the force in water reduces to

$\dfrac {1}{81} times$ .

A steady current is flowing in a cylindrical conductor.Does electric field exist within the conductor?

Yes,electric field exists within the conductor because it is the electric field which imparts acceleration to electrons for the flow of current.

In Fig. $22-35$ the four particles form a square of edge length $a=5.00\ cm$ and have charges $q_{1}=+10.0\ nC, q_{2}=-20.0\ nC, q_{3}=+20.0\ nC$ , and $q_{4}=-10.0\ nC$ , In unit vector notation, what net electric field do not particles produce at the square's center?

The

$x$ component of the electric field at the center of the square is given by

$E_{x}=\dfrac{1}{4\pi\varepsilon_{0}}\left[\dfrac{|q_{1}|}{(a/\sqrt{2})^{2}}+\dfrac{|q_{2}|}{(a/\sqrt{2})^{2}}-\dfrac{|q_{3}|}{(q/\sqrt{2})^{2}}-\dfrac{|q_{4}|}{(a/\sqrt{2})^{2}}\right]\cos 45^{o}$

$=\dfrac{1}{4\pi\varepsilon_{0}}\dfrac{1}{a^{2}/2}(|q_{1}|+|q_{2}|-|q_{3}|-|q_{4}|)\dfrac{1}{\sqrt{2}}$

$=0$ .
Similarly, the

$y$ component of the electric field is

$E_{y}=\dfrac{1}{4\pi\varepsilon_{0}}\left[-\dfrac{|q_{1}|}{(a/\sqrt{2})^{2}}+\dfrac{|q_{2}|}{(a/\sqrt{2})^{2}}+\dfrac{|q_{3}|}{(q/\sqrt{2})^{2}}-\dfrac{|q_{4}|}{(a/\sqrt{2})^{2}}\right]\cos 45^{o}$

$=\dfrac{1}{4\pi\varepsilon_{0}}\dfrac{1}{a^{2}/2}(-|q_{1}|+|q_{2}|+|q_{3}|-|q_{4}|)\dfrac{1}{\sqrt{2}}$

$=\dfrac{(8.99\times 10^{9}N.m^{2}/C^{2})(2.0\times 10^{-8}C)}{(0.050\ m)^{2}/2}\dfrac{1}{\sqrt{2}}=1.02\times 10^{5}N/C$
Thus, the electric field at the center of the square is

$\vec{E}=E_{y}\hat{j}=(1.02\times 10^{5}N/C)\hat{j}$ .

The net electric field is depicted in the figure below (not to scale). The field. pointing to the

$+y$ direction, is the vector sum of the electric fields of individual charges.

1697100_1771029_ans_56c67289b5ce4eb5b5823a4ede2f4f2b.png

Charge is uniformly distributed around a ring of radius $R=2.40\ cm$ , and the resulting field magnitude $E$ is measure along the ring's central axis (perpendicular to the plane of the ring). At what distance from the ring's centre is $E$ maximum?

We find the maximum by differentiating equation and setting the result equal to zero.

$\dfrac {d}{dz}\left(\dfrac {qz}{4\pi { \varepsilon }_{ 0 } (z^2 +R^2)^{3/2}}\right)=\dfrac {q}{4\pi { \varepsilon }_{ 0 }}$

$\dfrac {R^2 -2z^2}{4\pi { \varepsilon }_{ 0 }(z^2 +R^2)^{5/2}}=0$

which leads to

$z=R/\sqrt 2$ .

With

$R=2.40\ cm$ ,

we have

$z=1.70\ cm$ .

In the radioactive decay of Eq. $a^{238}U$ nucleus transforms to $^{234}Th$ and an ejected $^4He.$ (These are nuclei, not atoms, and thus electrons are not involved.)When the separation between $^{234}Th$ and $^4He$ is $9.0 \times 10-15^\ m$ , what are the magnitudes of the electrostatic force between then

We are concerned with the charges in the nucleus (not the "orbiting" electrons, if there are any). The nucleus of Helium has 2 protons and that of thorium has

$90$ .
Equation gives

$F=k\dfrac{q^2}{r^2}=\dfrac{(8.99 \times \ N.m^2/C^2)(2(1.60 \times 10^{-19\ }C))(90 (1.60 \times 10^{-19\ }C))}{(9.0 \times 10^{-15\ }m)^2}=5.1 \times 10^2\ N$

Figure shows three circular centred on the origin of a coordinate system. On each arc, the uniformly distributed charge is given in terms of $Q=2.00\ \mu C$ . The radii are given in term of $R=10.0\ cm$ .
What are the magnitude?

The smallest arc is of length

$L_1=\pi r_1 /2=\pi R/2$ ;

the middle-size arc has length

$L_2=\pi r_2/2=\pi (2\ R)/2$

$=\pi R$ ; and the large arc has

$L_3=\pi (3/R)/2$ .

The charge per unit length for each arc is

$\lambda =q/L$ where each charge

$q$ is specifics in the figure.

Thus, we find the net electric field to be

$E_{net}=\dfrac {\lambda_1(2\sin 45^o)}{4\pi { \varepsilon }_{ 0 } r_1}+\dfrac {\lambda_2 (2\sin 45^o)}{4\pi { \varepsilon }_{ 0 } r_2}+\dfrac {\lambda_3 (2\sin 45^o)}{4\pi { \varepsilon }_{ 0 } r_3}$

$=\dfrac {Q}{\sqrt 2 \pi^2 { \varepsilon }_{ 0 } R^2}$

which yield

$E_{net}-1.62\times 10^6 N/C$

The initial charges on the three identical metal spheres in Fig. are the following: sphere A, Q; sphere B, $-Q/4$ ; and sphere C, $Q/2$ , where $Q=2.00 \times 10^{-14} \ C$ . Spheres A and B are fixed in place, with a center-to-center separation of $d=1.20\ m$ , which is much larger than the spheres. Sphere C is touched first to sphere A and then to sphere B and is then removed. What then is the magnitude of the electrostatic force between spheres A and B?

When sphere C touches sphere A, they divide up their total charge

$(Q/2\ plus\ Q)$ equally between them. Thus, A now has charge

$3\ Q/4$ , and the magnitude of the force of attraction between A and B becomes

$F=k\dfrac{(3Q/4)(Q/4)}{d^2}=4.68 \times 10^{-19}\ N$ .

Is the force acting between two point charges $q1$ and $q2$ kept at some distance apart in air attractive or repulsive when (i) $q1q2 > 0$ (ii) $q1q2 < 0$ ?

(i) Repulsive

$q1q2 > 0$ suggests that both the charges are of the same polarity.

(ii) Attractive.

$q1q2 < 0$ suggests that both the charges are of the opposite polarity.

A $10.0\ g$ block with a charge of $+8.00 \times 10^{-5}C$ is placed in an electric field $\vec E=(3000 \hat i-600 \hat j)\ N/C$ . If the block is released from rest at the origin at time $t=0$ , what is its $y$ coordinate at $t=3.00\ s$ ?

With

$m=0.0100\ kg$ , the

$(x, y)$ coordinates at

$t=3.00\ s$ can be found

by combining Newton's second law with the kinematics equations of.

The

$x$ coordinate is

$x=\dfrac{1}{2}a_xt^2$

$=\dfrac{F_xt^2}{2m}$

$=\dfrac{(0.240\ N)(3.00\ s)^2}{2(0.0100\ kg)}=108\ m$ .

Similarly, the

$y$ coordinate is

$y=\dfrac{1}{2}a_yt^2=\dfrac{F_yt^2}{2m}=\dfrac{(-0.0480\ N)(3.00\ s)^2}{2(0.0100\ kg)}=-21.6\ m$ .

In figure, a nonconducting rod of length $L=8.15\ cm$ has a charge $-1=-4.23\ fC$ uniformly distributed along its length.
What are the direction (relative to the positive direction of the $x$ axis) of the electric field produced at point $P$ , at distance $a=12.0\ cm$ from the rod?

The negative sign in

$E_x$ indicates that

the field points in the

$-x$ direction,

$-180^o$ counterclockwise from the

$+x$ axis,

What is the nature of electrostatic force between two point electric charges $q_{1}$ and $q_{2}$ if
(a) $q_{1} + q_{2} > 0?$
(b) $q_{1} + q_{2} < 0?$

(a) If both

$q_{1}$ and

$q_{2}$ are positive, the electrostatic force between these will be repulsive.
However, if one of these charges is positive and is greater than the other negative charge, the electrostatic force between them will be attractive.
Thus, the nature of force between them can be repulsive or attractive.

(b) If both

$q_{1}$ and

$q_{2}$ are

$-ve$ , the force between these will be repulsive.
However, if one of them is

$-ve$ and it is greater in magnitude than the second

$+ve$ charge, the force between them will be attractive.
Thus, the nature of force between them can be repulsive or attractive.

A $10.0\ g$ block with a charge of $+8.00 \times 10^{-5}C$ is placed in an electric field $\vec E=(3000 \hat i-600 \hat j)\ N/C$ . If the block is released from rest at the origin at time $t=0$ , what is its $x$ coordinate at $t=3.00\ s$ ?

With

$m=0.0100\ kg$ , the

$(x, y)$ coordinates at

$t=3.00\ s$ can be found

by combining Newton's second law with the kinematics equations of.

The

$x$ coordinate is

$x=\dfrac{1}{2}a_xt^2$

$=\dfrac{F_xt^2}{2m}$

$=\dfrac{(0.240\ N)(3.00\ s)^2}{2(0.0100\ kg)}=108\ m$ .

In Fig the three particles are fixed in place and have charges $q_{1}=q_{2}=+e$ and $q_{3}=+2e$ . Distance $a=6.00 \mu m$ . What are the
direction of the net electric field at point $P$ due to the particles?

By symmetry we see that the contributions from the two charges

$q_{1}=q_{2}=+e$ cancel each other,

and we know that to compute magnitude of the field due to

$q_{3}=+2e$
The magnitude of the net electric field is

$|\vec{E_{net}}|=\dfrac{1}{4\pi\varepsilon_{0}}\dfrac{2e}{r^{2}}=\dfrac{1}{4\pi\varepsilon_{0}}\dfrac{2e}{(a/\sqrt{2})^{2}}=\dfrac{1}{4\pi\varepsilon_{0}}\dfrac{4e}{a^{2}}$

$=(8.99\times 10^{9}N.m^{2}/C^{2})\dfrac{4(1.60\times 10^{-19}C)}{(6.00\times 10^{-6}m)^{2}}$

$=160\ N/C$

This field points at

$45.0^{o}$ ,

counterclockwise from the

$x$ axis.

A $10.0\ g$ block with a charge of $+8.00 \times 10^{-5}C$ is placed in an electric field $\vec E=(3000 \hat i-600 \hat j)\ N/C$ . What is the magnitude.

We know.

$\vec F=(8.00\times 10^{-5}\ C)(3.00\times 10^3\ N/C)\hat i +(8.00\times 10^{-5}\ C)(-600\ N/C)\hat j$

$=(0.240\ N)\hat i-(0.0480\ N)\hat j$ .
Therefore, the force has magnitude equal to

$F=\sqrt{F_x^2+F_y^2}$

$=\sqrt{(0.240\ N)^2+(-0.0480\ N)^3}$

$=0.245\ N$

In figure, a nonconducting rod of length $L=8.15\ cm$ has a charge $-1=-4.23\ fC$ uniformly distributed along its length.
What are the magnitude ?

We position the

$x$ axis along the rod with the origin at the left and the rod, as shown in the diagram.
Let

$dx$ be an infinitesimal length of rod at

$x$ .

The charge on this segment is

$dq=\lambda \ dx$ .

The charge

$dp$ may be considered to be a point charge.

The electric field it produces at point

$P$ has only an

$x$ component, and this component is given by

$sE_x =\dfrac {1}{4\pi { \varepsilon }_{ 0 }}\dfrac {\lambda \ dx}{(L+a-x)^2}$
The total electric field product at

$P$ by the whole rod is the integral

$E_x =\dfrac {\lambda}{4\pi { \varepsilon }_{ 0 }}\displaystyle \int \dfrac {dx}{(L+a-x)^2} = \dfrac {\lambda}{4\pi { \varepsilon }_{ 0 }}\dfrac {1}{L+a-x}|_0^1 = \dfrac {\lambda }{4\pi { \varepsilon }_{ 0 }}\left(\dfrac {1}{a}-\dfrac {1}{L+a}\right)$

$=\dfrac {\lambda}{4\pi { \varepsilon }_{ 0 }}\dfrac {L}{a(L+a)}=-\dfrac {1}{4\pi { \varepsilon }_{ 0 }}\dfrac {q}{a(L+a)}$
upon substituting

$-q=\lambda L$ . With

$q=4.23\times 10^{15}C, L=0.0815\ m$ and

$a=0.120\ m$ , we obtain

$E_{x}=-1.57\times 10^{-3}N/ C$ ,

$|E_x|=1.57\times 10^{-3}N/C$ .

Five charges, $q$ each is placed at the corners of a regular pentagon of side $a$ as shown in figure. What will be the electric field at $O$ , the centre of the regular pentagon?

Point

$O$ is symmetric to all the five equal charges of

$+q$ placed at vertices of the regular polygon. So net electric field will be zero at

$O$ .

Can electric field at a point be zero, while electric potential is not zero

Yes, inside a hollow charged metallic conductor, the electric field is Zero, but electric potential is finite.

Five charges, $q$ each is placed at the corners of n-sided regular polygon of side $a$ . What will be the electric field at $O$ , the centre of the n-sided regular polygon? (See figure for analogy)

Electric field at

$O$ will be zero as all charge distribution about

$O$ is symmetric as in the case of the electric field at the centre of conducting ring or shell

Five charges, $q$ each is placed at the corners of a regular pentagon of side $a$ as shown in figure. What will be the electric field at $O$ if the charge from one of the corners (say $A$ ) is removed?

When a charge

$q$ from

$A$ is removed from symmetric charge distribution about

$O$ , the net electric field due to rest of the charges will be equal in magnitude as there is

$-q$ charge at

$A$ . So Electric field at

$O$ in this case

$E_{1}=\dfrac{-1q}{4\pi \epsilon_{0}r^{2}}$ ,

the direction of

$E$ is from

$O$ to

$A$

Five charges, $q$ each is placed at the corners of an n-sided regular polygon of side $a$ (see figure for analogy). What will be the electric field at $O$ if the charge $q$ at A is replaced by $-q$ ?

If charge

$-q$ is placed at one vertex after removing

$+q$ from there then resultant electric field at

$O$ will be due to charge (

$-2q$ ) at

$A$ .

So the net electric field at

$O$ after removing

$q$ from

$A$ by placing

$-q$ at

$A$ ,

$E=\dfrac{-q}{4\pi \epsilon_{0}r^{2}}-\dfrac{q}{4\pi \epsilon_{0}r^{2}}$

$E= \dfrac{-2q}{4\pi \epsilon_{0}r^{2}}$ from

$O$ to

$A$ .

For two point charges and $q_1$ and $q_2$ , $q_1q_2 > 0$ . What is the nature of electrostatic force between two charges?

$q_1q_2 > 0$ , then both

$q_1$ and

$q_2$ are likely charged (either both positively charged or negatively charged) and there will be the electrostatic force of repulsion between them.

The electrostatic force between two protons placed at distance r is F. If we keep electrons in place of protons, then what would be the effect on the electrostatic force?

Charge on proton = Charge on electron

∴ There will be no change in the electrostatic force.

∴ Electrostatic force = F

For two point charges $q_1$ and $q_2$ , $q_1q_2<0$ . What is the nature of electrostatic force between two charges?

$q_1q_2 < 0$ , then either of

$q_1$ or

$q_2$ is positively charged and the other is negatively charged and there is the force of attraction between them.

The electrostatic force (F) acts between two point charges in a vacuum. If a brass plate is placed between the two charges. What would be the value of the electrostatic force?

Force acting between two-point charge in vacuum

$F =\dfrac{1}{4\pi \epsilon_0}\,\dfrac{q_1q_2}{r^2}$ force between point charge in medium

$F_m =\dfrac{1}{4\pi \epsilon}\,\dfrac{q_1q_2}{r^2}=\dfrac{1}{4\pi \epsilon_0\epsilon_r}\,\dfrac{q_1q_2}{r^2}$

$\therefore F_m=\dfrac{F}{\epsilon_r}=\dfrac{F}{\infty}=0$

The electric potential at a point $(x, y)$ in an electric field is given by $V=6 x y+y^{2}-x^{2}$ Calculate the electric field at that point.

$\begin{array}{c}\vec{E}=-\vec{\nabla} V, \vec{E}=-\dfrac{\partial V}{\partial x} \hat{i}-\dfrac{\partial V}{\partial y} \hat{j}-\dfrac{\partial V}{\partial z} \hat{k} \\\vec{V}=\left[6 x y+y^{2}-x^{2}\right] \\\vec{E}=-\dfrac{\partial}{\partial x}\left(6 x y+y^{2}-x^{2}\right) \hat{i}-\dfrac{\partial}{\partial y}\left[6 x y+y^{2}-x^{2}\right] \dot{j} -\dfrac{\partial}{\partial z}\left[6 x y+y^{2}-x^{2}\right] \hat{k}\\\vec{E}=-[6 y-2 x] \hat{i}-[6 x+2 y] \hat{j}-0 \\\vec{E}=(2 x-6 y) \hat{i}-(6 x+2 y) \hat{j \ } \mathrm{volt / m} \end{array}$

Two point charges $q_A=3\ \mu C$ and $q_B=3\ \mu C$ are located $20\ cm$ apart in vacuum.
If a negative test charge of magnitude $1.5\times 10^{-9}C$ is placed at this point, what is the force experienced by the test charge?

Here,

$q_A=3\ \mu C=3\times 10^{-6}C$

$q_B=-3\ \mu C=-3\times 10^{-6}C$

Let

$O$ be the midpoint of the line

$AB$

then

$OA=OB=\dfrac r2 =\dfrac {0.2}{2}=0.1\ m$ .

Force on a negative charge of magnitude

$1.5\times 10^{-9}C$ placed at point

$O$ ,

$F=qE$

$=1.5\times 10^{-9}C\times 5.4\times 10^6 NC^{-1}$

$=8.1\times 10^{-3}N$ (along

$OA$ )

The force on the negative charge acts in a direction opposite to that of the electric field.

What is the force between two small charged spheres having charges of $2\times 10^{-7}$ and $3\times 10^{-7} C$ placed $30\ cm$ apart in air?

$q_1=2\times 10^{-7}C$

$q_2=3\times 10^{-7}C$

$r=30\ cm =0.3\ m$

$F=\dfrac {1}{4\pi \in_0}\times \dfrac {q_1 q_2}{r^2}$

$=9\times 10^9 \times \dfrac {2\times 10^{-7}\times 3\times 10^{-7}}{(0.3)^2}$

$=6\times 10^{-3}N$

A non-polar molecule with polarizability $\beta$ is located at a great distance $l$ from a polar molecule with electric moment $p$ . Find the magnitude of the interaction force between the molecules if the vector $p$ is oriented along a straight line passing through both molecules.

From the general formula

$\vec {E} = \dfrac {1}{4\pi \epsilon_{0}} \dfrac {3 \vec {p}\cdot \vec {r}\vec {r} - \vec {p}r^{2}}{r^{5}}$

$\vec {E} = \dfrac {1}{4\pi \epsilon_{0}} \dfrac {2\vec {p}}{l^{3}}$ , where

$r = l$ and

$\vec {r} \uparrow \uparrow \vec {p}$
This will cause the induction of a dipole moment.

$\vec {p}_{ind} = \beta \dfrac {1}{4\pi \epsilon_{0}} \dfrac {2\vec {p}}{l^{3}} \times \epsilon_{0}$
Thus the force,

$\vec {F} = \dfrac {\beta}{4\pi} \dfrac {2p}{l^{3}} \dfrac {\partial}{\partial l} \dfrac {1}{4\pi \epsilon_{0}} \dfrac {2p}{l^{3}} = \dfrac {3\beta p^{2}}{4\pi^{2} \epsilon_{0}l^{7}}$ .

Two point charges, $q$ and $-q$ , are separated by a distance $l$ , both being located at a distance $l/2$ from the infinite conducting plane. Find:
(a) the modulus of the vector of the electric force acting on each charge;
(b) the magnitude of the electric field strength vector at the mid-point between these charges.

(a) Using the concept of electrical image, it is clear that the magnitude of the force acting on each charge,

$|\vec {F}| = \sqrt {2} \dfrac {q^{2}}{4\pi \epsilon_{0}l^{2}} - \dfrac {q^{2}}{4\pi \epsilon_{0}(\sqrt {2}l)^{2}}$

$= \dfrac {q^{2}}{8\pi \epsilon_{0}l^{2}} (2\sqrt {2} - 1)$

(b) Also, from the figure, magnitude of electrical field strength at

$P$

$E = 2\left (1 - \dfrac {1}{5\sqrt {5}}\right ) \dfrac {q}{\pi \epsilon_{0} l^{2}}$ .

1784021_1851209_ans_09987337f93f416eb55774b4736887da.png

Consider two electric dipoles in empty space. Each dipole has zero net charge.
(a) Does an electric force exist between the dipoles; that is, can two objects with zero net charge exert electric forces on each other?
(b) If so, is the force one of attraction or of repulsion?

(a) Yes.
(b) The situation is similar to that of magnetic bar magnets, which can attract or repel each other depending on their orientation.

The electrons in a particle beam each have a kinetic energy $K$ . What are (a) the magnitude and (b) the direction of the electric field that will stop these electrons in a distance $d$ ?

The work done on the charge is

$W=\overrightarrow{F}\cdot\overrightarrow{d}=q\overrightarrow{E}\cdot\overrightarrow{d}$ and the kinetic energy changes according to

$W=K_f-K_i=0-K$ .
Assuming

$\overrightarrow{v}$ is in the

$+x$ direction, we have

$(-e)\overrightarrow{E}\cdot d\hat{i}=-K$ .
Then,

$e\overrightarrow{E}\cdot(d\hat{i})=K$ , and

$\overrightarrow{E}=\dfrac{K}{ed}\hat{i}$
(a)

$E=\boxed{\dfrac{K}{ed}}$
(b) Because a negative charge experiences an electric force opposite to the direction of an electric field, the required electric field will be

$\boxed{\text{in the direction of motion}}$ .

A thin insulator rod is placed between two unlike point charges $+{q}_{1}$ and $-{q}_{2}$ (figure).
How will the forces acting on the charges change?

Due to polarization of the insulator rod

$AB$ the point charge

$+{q}_{1}$ will be acted upon in addition to thepoint charge

$-{q}_{2}$ , by the polarization charges formed at the ends of the rod (figure).
The attractive force exerted by the negative charge induced at the end

$A$ will be stronger than the repulsive force exerted by the positive charge induced at the end

$B$ . Thus, the total force acting on the charge

$+{q}_{1}$ will increase,
1806610_1860059_ans_9db8cf12f97b46e08339f89cad1c3906.png

Coulomb's Law
A $7.50-nC$ point charge is located $1.80 \,m$ from a $4.20-nC$ point charge.
(a) Find the magnitude of the electric force that one particle exerts on the other.
(b) Is the force attractive or repulsive?

(a)

$|F|=\dfrac{k_e|q_1||q_2|}{r^2}$

$F=\dfrac{k_e e^2}{r^2}=\dfrac{(8.99\times 10^{9} N.m^2/C^2)(7.50\times 10^{-9} C)(4.20\times 10^{-9}C)}{(1.80 \,m)^2}$

$=\boxed{8.74\times 10^{-8} \,N}$
(b) The charges are like charges.

$\boxed{\text{The force is repulsive.}}$

Coulomb's Law
(a) Find the magnitude of the electric force between a $Na^{+}$ ion and a $Cl^{-}$ ion separated by $0.50 \,nm$ .
(b) Would the answer change if the sodium ion were replaced by $Li^{+}$ and the chloride ion by $Br^{-}$ ? Explain.

(a) The two ions are both singly charged,

$|q|=1e$ , one positive and one negative. Thus,

$|F|=\dfrac{k_e|q_1||q_2|}{r^2}=\dfrac{k_e e^2}{r^2}$

$=\dfrac{(8.99\times 10^{9} N.m^2/C^2)(1.60\times 10^{-19} C)^2}{(0.500\times 10^{-9} \,m)^2}$

$=\boxed{9.21\times 10^{-10} N}$
(b) No. The electric force depends only on the magnitudes of the two charges and the distance between them.

Consider a plane surface in a uniform electric field as in Figure, where $d=15.0 cm$ and $\theta=70.0^{o}$ . If the net flux through the surface is $6.00 N$ ? $m^{2}/C$ , find the magnitude of the electric field.

The electric field makes an angle of

$70.0^{o}$ to the normal. The square has side

$d=5.00cm$

$\phi_{E}=EA\cos \theta=Ed^{2}\cos \theta$

$\rightarrow E=\dfrac{\phi_{E}}{d^{2}\cos \theta}=\dfrac{6.00N\cdot m^{2}/C}{(0.150m)^{2}\cos 70.0^{o}}=780N/C$

Two identical metal spheres having equal and similar charges repel each other with a force of 103 N when they are placed 10 cm apart in a medium of dielectric constantDetermine the charge on each sphere.

Step-2 After putting the given values.

Two pith balls carrying equal charges are suspended from a common point by strings of equal length, the equilibrium separation between them is $r$ . Fig. Now the strings are rigidly clamped at half the height. The equilibrium separation between the balls, now becomes.

Given two pith balls carrying equal charges are suspended from a common point by strings of equal length. The equilibrium separation between them is

$r$ . Now, we have to find the equilibrium separation when the strings are clamped at half the height.

We can see from the figure that under equilibrium,

$\tan{\theta}=\dfrac{F}{mg}$ ...(i)

and also,

$\tan{\theta}=\dfrac{r/2}{y}$ ...(ii)

Equating (i) and (ii), we get

$\dfrac{F}{mg}=\dfrac{r}{2y}$

also, Force,

$F=\dfrac{kq^{2}}{r^2}$

where

$k=$ constant

$=9\times 10^{9}$

So, we get

$r^{3}=\left ( \dfrac{kq^{2}}{mg} \right )y$ ...(iii)

Let

$r'$ be the equilibrium separation when

$y\rightarrow \dfrac{y}{2}$

So, equation (iii) then reduces to:

$r'^{3}=\left ( \dfrac{kq^{2}}{mg} \right )\dfrac{y}{2}$ ...(iv)

Dividing (iv) by (iii) we get,

So,

$\dfrac{r'^{3}}{r^{3}}=\dfrac{1}{2}$

$r'=r\left ( \dfrac{1}{2} \right )^{1/3}$

Two small spheres are charged positively, the combined charge being $5.0\times { 10 }^{ -5 }C$ . If each sphere is repelled from the other by a force of $1.0N$ when the spheres are $2.0$ metre apart, how is the total charge distributed between the spheres?

Let the charge on one sphere be

$q_1$ .

Charge on other sphere

$q_2 = 5\times 10^{-5} - q_1$

Force between them

$F =\dfrac{Kq_1q_2}{d^2}$

where

$d = 2 \ m$

Or,

$1 = \dfrac{9\times 10^9\times q_1 (5\times 10^{-5} -q_1)}{2^2}$

$\implies \ q_1^2 - 5\times 10^{-5}q_1 + 0.44 \times 10^{-9}= 0$

We get

$q_1 = 1.15\times 10^{-5} \ C$

Charge on other sphere

$q_2 = (5-1.15)\times 10^{-5} = 3.85\times 10^{-5} \ C$

At what point and what charge should be placed so that the entire system is in equilibrium?

At point O, there is charge by which there is equilibrium,

$\therefore \quad { \vec { E } }_{ OA }={ \vec { E } }_{ OB }$

$\Rightarrow \quad \dfrac { K\left( Q \right) }{ { x }^{ 2 } } =\dfrac { K\left( Q \right) }{ { \left( d-x \right) }^{ 2 } }$

$\Rightarrow \quad { x }^{ 2 }={ d }^{ 2 }-2dx+{ x }^{ 2 }$

${ d }^{ 2 }=2dx$

$\boxed { x=\dfrac { d }{ 2 } }$

$d/2$ there should be charge by which there will be equilibrium.

For, magnitude,

${ \vec { F } }_{ OA }=\dfrac { K\left( Q \right) \left\{ x \right\} }{ { \left( d/2 \right) }^{ 2 } }$

where

$K$ the charge present at

$d/2$ .

Also,

${ \vec { F } }_{ BA }=\dfrac { KQ\left( Q \right) }{ { d }^{ 2 } }$

$\therefore$ For equilibrium,

${ \vec { F } }_{ OA }+{ \vec { F } }_{ BA }=0$

$\Rightarrow \quad \dfrac { KQx }{ { \left( d/2 \right) }^{ 2 } } =\dfrac { { -KQ }^{ 2 } }{ { d }^{ 2 } }$

$\Rightarrow \quad x=\dfrac { -Q }{ 4 }$

$\therefore$ At

$(d/2)$ a charge of

$-Q/4$ should be placed for the system to be in equilibrium.

The diagram shows the arrangement of three small uniformly charged spheres $A,B$ and $C.$ The arrows indicate the direction of the electrostatic forces acting between the spheres (for example, the left arrow on sphere $A$ indicates the electrostatic force on sphere $A$ due to sphere $B$ ). At least two of the spheres are positively charged. Which sphere, if any, could be negatively charged?

$A{\text{ }}is{\text{ }}\left( + \right)ve$

$B{\text{ }}\& {\text{ }}C{\text{ }}\left( - \right)ve$

Two particles $A$ and $B$ , each having a charge $Q$ , are placed a distance $d$ apart. Where should a particle of charge $q$ be placed on the perpendicular bisector of $AB$ so that it experiences maximum force ? What is the magnitude of this maximum force ?

Hint:

Magnitude is described as big in length or very important.
An instance of significance is the intensity of the Grand Canyon.
An instance of significance is the dimensions of the hassle of globalwide hunger.

Step 1 : Find the force of the charge particle.

Let,

Force on the charge particle

$‘q’$ at

$‘c’$ is only the x component of

$2$ forces

So,

$\Rightarrow$

$F_{\text {on } c}=F_{C B} \operatorname{Sin} \theta+F_{A C} \operatorname{Sin} \theta$

But

$\left|\bar{F}_{\text {CB }}\right|=\left|\bar{F}_{A C}\right|$

$\Rightarrow2F_{CB}\operatorname{Sin}\theta$

$\Rightarrow2\frac{KQq}{x^2+(d/2)^2}\times\frac{x}{\left[x^2+d^2/4\right]^{1/2}}$

$\Rightarrow\frac{2k\theta qx}{\left(x^2+d^2/4\right)^{3/2}}$

$\Rightarrow\frac{16kQq}{\left(4x^2+d^2\right)^{3/2}}x$

Step 2 : Find the magnitude of maximum force.

For maximum force

$\frac{d F}{d x}=0$

$\Rightarrow$

$\frac{d}{d x}\left(\frac{16 k Q q x}{\left(4 x^{2}+d^{2}\right)^{3 / 2}}\right)=0$

$\Rightarrow$

$K\left[\frac{\left(4 x^{2}+d^{2}\right)-x\left[3 / 2\left[4 x^{2}+d^{2}\right]^{1 / 2} 8 x\right]}{\left[4 x^{2}+d^{2}\right]^{3}}\right]=0$

$\Rightarrow \frac{\mathrm{K}\left(4 \mathrm{x}^{2}+\mathrm{d}^{2}\right)^{1 / 2} \mid\left(4 \mathrm{x}^{2}+\mathrm{d}^{2}\right)^{3}-12 \mathrm{x}^{2}}{\left(4 \mathrm{x}^{2}+\mathrm{d}^{2}\right)^{3}}=0$

$\Rightarrow\left(4 \mathrm{x}^{2}+\mathrm{d}^{2}\right)^{3}=12 \mathrm{x}^{2}$

$\Rightarrow 16 x^{4}+d^{4}+8 x^{2} d^{2}=12 x^{2}$

$\Rightarrow$

$\quad d^{4}+8 x^{2} d^{2}=0$

$\Rightarrow d^{2}=0 \quad d^{2}+8 x^{2}=0$

$\Rightarrow d^{2}=8 x^{2} \Rightarrow d=\frac{d}{2 \sqrt{2}}$ .

Therefore, The magnitude of this maximum force is

$\frac{d}{2 \sqrt{2}}$ .

Two particles $A$ and $B$ having charges $8 \times 10^{-6}C$ and $-2 \times 10^{-6}$ respectively are held fixed with a separation of $20 cm$ . Where should a third charged particle $C$ be placed so that it does not experiences a net electric force?

As the net electric force

$C$ should be equal to zero, the force due to

$A$ and

$B$ must be opposite in direction. Hence, the particle should be placed on the line

$AB$ . As

$A$ and

$B$ have changes of opposite signs,

$C$ cannot be between

$A$ and

$B$ .
Also

$A$ has larger magnitude of charge than

$B$ . Hence,

$C$ should be placed closer to

$B$ that

$A$ . The situation is shown in figure. Suppose

$BC = x$ and the charge on

$C$ is

$Q$

$\overrightarrow{F}_{CA} = \dfrac{1}{4\pi \in_0}\dfrac{(8.0 \times 10^{-6})Q}{(0.2 + x)^2}\hat{i}$

and

$\overrightarrow{F}_{CB} = \dfrac{-1}{4\pi \in_0}\dfrac{(2.0 \times 10^{-6})Q}{x^2}\hat{i}$

$\overrightarrow{F}_C = \overrightarrow{F}_{CA} + \overrightarrow{F}_{CB} = \dfrac{1}{4\pi \in_0}\left[\dfrac{(8.0 \times 10^{-6})Q}{(0.2 \times 10^{-6})} - \dfrac{(2.0 \times 10^{-6})Q}{x^2}\right]\hat{i}$
But

$|\overrightarrow{F}_C| = 0$

Hence

$\dfrac{1}{4\pi \in_0}\left[\dfrac{(8.0 \times 10^{-6})Q}{(0.2 \times x)^2} - \dfrac{(2.0 \times 10^{-6})Q}{x^2}\right] = 0$
Which gives

$x = 0.2 m$
1029653_1013491_ans_ae2c8574ebe84343ae162d9825a3f68c.png

Two positively charged particles, each of mass 1.7 x $10^{-27}$ kg and carrying a charge of 1.6 x $10^{-19}$ C are placed at a distance d apart. If each experiences a repulsive force equal to its weight, find the value of d.

Given :

$m = 1.7\times 10^{-27} \ kg$

$q = 1.6\times 10^{-19} \ C$

$F_{electrostatic} = F_{gravity}$

We have

$\dfrac{kq^2}{d^2} = mg$

$\dfrac{9\times 10^9 (1.6\times 10^{-19})^2}{d^2} = 1.7\times 10^{-27}\times 10$

$\implies \ d = 0.11 \ m = 11 \ cm$

The distance between electron and proton in the H-atom is about $5.3 \times 10^{-11}$ metre. What is the magnitude of electric force between these two particles?
$\left( \dfrac {1}{4 \pi \epsilon_0} = 9 \times 10^9 \dfrac {Nm^2}{C^2} \right)$

The electric force between the particles is given as,

$F = \dfrac{1}{{4\pi {\varepsilon _0}}} \times \dfrac{{{Q_1}{Q_2}}}{{{r^2}}}$

$F = 9 \times {10^9} \times \dfrac{{1.6 \times {{10}^{ - 19}} \times 1.6 \times {{10}^{ - 19}}}}{{{{\left( {5.3 \times {{10}^{ - 11}}} \right)}^2}}}$

$F = 0.82 \times 10{}^{ - 7}\;{\rm{N}}$

Three equal changes $2.0 \times 10 ^ {-6} C$ each, are held fixed at the three corners of an equilateral triangle of side 5 cm. Find the Coulomb force experienced by one of the charges due to the rest two.

$Q = 2\times 10^{-6}C$

To find-

Coulomb force experienced by one of the charge due to other two changes

$|\vec{F_1}| = \dfrac{Kq^2}{a^2} = F$

$|\vec{F_2}| = \dfrac{Kq^2}{a^2} = F$

$F_{Net} = \sqrt{F_!^2 + F_2^2+2F_1F_2\cos \theta}$

$=\sqrt{F^2+F^2+2F^2\cos 60}$

$= \sqrt{3}F$

$=\sqrt{3}\times \dfrac{Kq^2}{a^2}$

$= \dfrac{\sqrt{3}\times 9 \times 10^9\times (2\times 10^{-6})^2}{(5\times 10^{-2})^2}$

$=\dfrac{\sqrt{3}\times 9\times 10^9\times 4\times 10^{-12}}{25\times 10^{-4}}$

$= \dfrac{\sqrt{3} \times 9\times 4\times 10}{25}$

$= 24.94N$

1461766_1039562_ans_b6487bde29b3470382333882f667b84e.png

A total change Q is broken in two parts $Q_1$ and $Q_2$ and they are placed at a distance R from each other. The maximum force of repulsion between them will occur, when

$Q_2=Q-Q_1$

$F=\dfrac{k}{R^2}Q_1Q_2=\dfrac{k}{R^2}Q_1(Q-Q_1)$

$\implies F=\dfrac{k}{R^2}(Q_1Q-Q_1^2)$

Differentiating w.r.t

$Q_1$ and equating to

$0$ for maximum force-

$Q-2Q_1=0$

$\implies Q_1=\dfrac{Q}{2}$

Hence, maximum repulsion will be there when both charges are equal and have

$Q_1=Q_2=\dfrac{Q}{2}$

Charge q is divided into two part and is placed at some fixed distance apart. What should be the value of the two parts such that force between them is maximum?

Let two parts be

$Q$ and

$(q-Q)$

$F=\dfrac{kQ(q-Q)}{r^2}=\dfrac{k}{r^2}(qQ-Q^2)$

For maximizing the force, differentiating both sides w.r.t

$Q$ and equating ot

$0$ .

$\dfrac{dF}{dQ}=\dfrac{k}{r^2}(q-2Q)=0$

$\implies Q=\dfrac{q}{2}$

Hence, the two parts of charge should have equal charge each equal to

$\dfrac{q}{2}$

N fundamental charge each of charge 'q' are to be distributed as two point charges seperated by a fixed distance then the maximum to minimum force bears a ratio ( N is even and greater than 2 )

Let charges be

$Q_1=nq$ and

$Q_2=(N-n)q$

$F=\dfrac{kQ_1Q_2}{r^2}$

$F=\dfrac{kn(N-n)q^2}{r^2}=\dfrac{kq^2}{r^2}n(N-n)$

For maximum force, the differential of

$F$ w.r.t

$n$ should be

$0$ .

$\implies N-2n=0$

$\implies n=\dfrac{N}{2}$

Hence,

$Q_1=Q_2=\dfrac{Nq}{2}$

Two charges $2 \cdot 0 \times 10 ^ { - 6 } \mathrm { C }$ and $1.0 \times 10 ^ { - 6 } \mathrm { C }$ are placed at a separation of $10 cm$ . Where should a third charge be placed such that it experiences no net force due to these charges?

If booth are positive charge then the third charge should be

placed on the line joining the two charge.

if should be in between the two charge

$F_1 = F _2$

$\dfrac{2 \times q}{r^2}$ = $\dfrac{1 \times q}{( 10 - r )^2}$

$r^2 = 2 ( 10 - r )^2$

$r^2 = 200 + 2 r^2 - 40r$

$r^2 - 40r + 200 = 0$

$r = \dfrac{[ 40 _{-}^{+} \sqrt(1600 - 800 )]}{2}$

= $\dfrac{[40 - \sqrt(1600 - 800 )]}{2}$

since the point should be in between

= $20 - 10 \sqrt2$

= 5.86 cm

third charge should be placed at 5.86 cm

from $2 \times 10^{-6}$ C charge.

Two neutral particle like identical metal balls each of radius r and mass m are connected by a light inextensible conducting thread of length L. The balls are held in free space with separation between their centres l (r << l < L). Now everywhere a uniform electric field directed along the line joining the centres of the balls is switched on and then the balls are released. Find maximum speeds of the balls in subsequent motion.

1674480_1732015_ans_97b0a1c92dbc4c418d7fec3a8bb8d88f.jpg

A circular wire-loop of radius $a$ carries a total charge Q distributed uniformly over its length. A small length $dl$ of the wire is cut off. Find the electric field at the centre due to the remaining wire.

$\textbf{Step 1: Breaking the Loop in two parts}$

Let's divide the loop in two parts.

$1.$ The Small Element of length

$dl$ , which is to be cut off

$2.$ The Remaining part of the loop

Initially, by symmetry, the net Electric field at the center is zero.

$\therefore\ \ E_{Remaining\ Wire} +E_{dl\ Element} = 0$

$\Rightarrow\ \ E_{Remaining\ Wire} = -E_{dl\ Element}$

$....(1)$

$\textbf{Step 2: Electric field of Small Element dl}$

The charge per unit length of the wire is

$=\dfrac{Q}{2\pi a}=\lambda$

The Charge on the

$dl$ length of wire,

$dQ=\lambda\ dl=\dfrac{Qdl}{2\pi a}$

The small element can be considered as point charge, So its Electric Field at the centre:

$E_{dl\ Element} = \cfrac{k\ dQ}{a^2}$

$=\cfrac{1}{4\pi\ \epsilon_0}\times \cfrac{Q\ dl}{2\pi\ a^3}$

$=\dfrac{Qdl}{8\pi^2\varepsilon_0a^3}$

$....(2)$

$\textbf{Step 3: Electric field of Remaining wire}$

Hence, using Equation

$(1)$ and

$(2)$ , we get:

$\mid E_{Remaining\ Wire} \mid =$

$\dfrac{Qdl}{8\pi^2\varepsilon_0a^3}$

2112179_1720233_ans_675fdc3d2cfc4abcba6f0b11bb3212b6.png

A block of mass m having a charge q is placed on a smooth horizontal table and is connected to a wall through an unstressed spring of spring constant k as shown in figure. A horizontal electric field E parallel to the spring is switched on. Find the amplitude of the resulting SHM of the block.

The force due to the electric field on the mass is given as:

$F=qE$

The restoring force developed in the spring is given by:

$F=-Kx$

Where

$x=$ amplitude

At maximum elongation (i.e. amplitude), the two forces will balance each other.

$qE=-Kx$

$x=\dfrac{-qE}{K}$

1564281_1720362_ans_6c1c9b841ebf45e89b2930b38bdcf0e0.png

A particle of mass m and charge q projected with a speed u at an angle $\theta$ above the horizontal, after travelling a horizontal distance $x_0$ enters a region where in addition to gravity a uniform static horizontal electric field also exists. Boundary of this region is vertical. If after some time, the particle returns to the point of projection, find magnitude of the electric field and time of flight of the particle. Acceleration of free fall is g and influence of the air is to be neglected.

1657604_1731875_ans_1817a62dc3544c3786665201187842ef.jpg

In figure particle $1$ of charge $+4e$ is above a floor by distance $d_1=2.00\ mm$ and particle $2$ of charge $+6e$ is above a floor by distance $d_2=6.00\ mm$ horizontally from particle $1$ . What is the x component of the electrostatic force on particle $2$ due to particle $1$ ?

Using Coulomb's law, the magnitude of the force of particle

$1$ on particle

$2$ is

$f_21=k\dfrac{q_1q_2}{r^2}$ , where

$r=\sqrt{d_1^2+d_1^2}$ and

$k=1/4\pi\varepsilon_0=8.99 \times 10^9N.m^2/C^2$ .

Since both

$q_1$ and

$q_2$ are positively charged, particle

$2$ is repelled by particle

$1$ , so the direction of

$\vec { F } _{21}$ is away from particle

$1$ and toward

$2$ . In unit-vector notation,

$\vec F_{21}=\vec F_{21}\hat { r }$ ,

where

$\hat { r }=\dfrac{\vec r}{r}=\dfrac{(d_2 \hat i-d_1\hat j)}{\sqrt{d_1^2+d_2^2}}$ The

$x$ component of

$\vec { F } _{21}$ is

${ F } _{21 x}=F_{21}d_2 / {\sqrt{d_1^2+d_2^2}}$ . Combining the expressions above, we obtain

${ F } _{21 x}=k\dfrac{q_1q_2d_2}{r^3}=k\dfrac{q_1q_2d_2}{{d_1^2+d_2^2}}{3/2}$

$=\dfrac { { (8.99\times 10^{ 9 }N.m^{ 2 }/C^{ 2 })(4.1.60\times 10^{ -19 }C)() }{ 6.4.60\times 10^{ -19 }C(6.00\times 10^{ -3 }m) } }{ { [(2.00\times 10^{ -3 }m)^{ 2 }+(6.00\times 10^{ -3 }m)^{ 2 } }]^{ 3/2 } }$

$=1.31 \times 10^{-22} N$ Note:

In a similar manner,

we find the

$y$ component of

${\vec F_{21}}$ to be

${ F } _{21 y}=-k\dfrac{q_1q_2d_2}{r^3}=-k\dfrac{q_1q_2d_2}{{d_1^2+d_2^2}}${3/2}$

$=-0.437 \times 10^{-22} N$ Thus,

$\vec F_{21}=(1.31 \times 10^{-22} N)\hat i - (0.437 \times 10^{-22} N) \hat j$ .

An electron enters a region of uniform electric field with an initial velocity of $40\ km/s$ in the same direction as the electric field, which has magnitude $E=50\ N/C$ .
How far does the electron travel during the $1.5\ ns$ interval?

Given:

Electric field

The direction of the electric field and the direction of the velocity of the electron are the same. Therefore,

Of the charge $Q$ initially on a tiny sphere, $a$ portion $q$ is to be transferred to a second, nearby sphere. Both spheres can be treated as particles and are fixed with a certain separation. For what value of $q/Q$ will the electrostatic force between the two sphere be maximized?

The magnitude of the force of either of the charge on the other is given by

$F=\dfrac {1}{4\pi { \varepsilon }_{ 0 }}\dfrac {q(Q-q)}{r^2}$
where

$r$ is the distance between the charges. We want the value of

$q$ that maximizes the function

$f(q)=q(Q-q)$ . Setting derivative

$\dfrac{dF}{dp}$ equal to zero leads to

$Q-2q=0$ or

$q=Q/2$ .

Thus

$q/Q=0.500$ .

Two engineering students, John with a mass of $90\ kg$ and Mary with a mass of $45\ kg$ , are $30\ m$ apart. Suppose each has a $0.01 \%$ imbalance in the amount of positive and negative charge, one student being positive and the other negative. Find the order of magnitude of the electrostatic force of attraction between them by replacing each student with a sphere of water having the same mass as the student.

The net charge carried by John whose mass is m roughly

$q=(0.0001)\dfrac{mN_AZe}{M}$

$=(0.0001)\dfrac{(90\ kg)(6.02 \times 10^{23}\ molecules/mol)(18\ electron\ proton\ pairs/molecule)(1.6 \times 10^{-19}C)}{0.018\ kg/mol}$

$=8.7 \times 10^5\ C$ and the net charge carried by Mary is half of that. So the electrostatic force between the miss estimated to be

$F=k\dfrac{q(q/2)}{d^2}=(8.99 \times 10^8\ N.m^2/C^2)\dfrac{(8.7 \times 10^5 C)^2}{2(30m)^2} \approx 4 \times 10^{18}\ N$ .
Thus, the order of magnitude of the electrostatic force is

$10^{18}\ N$ .

In figure, three identical conducing spheres initially have the following charges: sphere $A, 4Q$ ; sphere $B, -6Q$ ; and sphere $C, 0$ . Sphere $A$ and $B$ are fixed in place, with a center-to-center separation that is much larger than the spheres. Two experiments are conduced. In experiment $1$ , sphere $C$ is touched to sphere $A$ and then (separately) to sphere $B$ , and then it removed. In experiment $2$ , starting with the same initial state, the procedure is reversed: Sphere $C$ is touched to sphere $B$ and then to sphere $A$ , and then it is removed. What is the ratio of the electrostatic force between $A$ and $B$ at the end of experiment $2$ to that at the end of experiment $1$ ?

In experiment

$1$ , sphere

$C$ first touches sphere,

$A$ , and they divided up their total charge (

$Q/2$ plus

$Q$ ) equally between them. Thus, sphere

$A$ and sphere

$C$ each acquired charge

$3Q/4$ . Then, sphere

$C$ touches

$B$ and those spheres split up their total charge

$(3Q/4)$ plus

$(Q/4)$ so that

$B$ ends up with charge equal to

$Q/4$ . The force of repulsion between

$A$ and

$B$ is therefore

$F_1=k\dfrac {(3Q/4)(Q/4)}{d^2}$
at the end of experiment

$1$ . Now, in experiment

$2$ , sphere

$C$ first touches

$B$ , which leaves each of them charge

$Q/8$ . When

$C$ next touches

$A$ is left with charge

$9Q/16$ . Consequently, the force of repulsion the force of repulsion between

$A$ and

$B$ is

$F_2=k\dfrac {(9Q/16)(Q/8)}{d^2}$
at the end experiment

$2$ . The ratio is

$\dfrac {F_2}{F_1}=\dfrac {(9/16)(1/8)}{(3/4)(1/4)}=0.375$ .

Identical isolated conducing sphere $1$ and $2$ have equal charge and are separated by a distance that is large compared with their diameters. The electrostatic force acting on sphere $2$ due to sphere $1$ is $\vec F$ . Suppose now that a third identical sphere $3$ , having an insulting handle and initially neutral, is touches first to sphere $1$ , then to sphere $2$ and finally removed. The electrostatic force that now acts on sphere $2$ has magnitude $F'$ . What is the ratio $F'/F$ ?

The fact that the sphere is identical allows us to conclude that when two-sphere are in contact, they share charge. Therefore, when a charges sphere

$(q)$ touches an unchanged one, they will (fairly quickly) each attain half that charge

$(q/2)$ . We start with spheres

$1$ and

$2$ , each having charge

$q$ and experiencing a mutual repulsive force

$F=kq^2/r^2$ . When the neutral sphere

$3$ touches

$1$ , sphere

$1\ s$ charge decreases to

$q/2$ . Then sphere

$3$ is brought into contact with sphere

$2; a$ total amount of

$q/2+q$ becomes equally between them. Therefore, the charge of sphere

$3$ is

$3q/4$ in the final situation. The repulsive force between sphere

$1$ and

$2$ is finally

$F=k \dfrac {(q/2)(3q/4)}{r^2}=\dfrac {3}{8}k\dfrac {q^2}{r^2}=\dfrac {3}{8}F\Rightarrow \dfrac {F'}{F}=0.375$

The variation of electric potential for an electric field directed parallel to the x-axis is shown in the given graph. Draw the variation of electric field strength with x-axis.

The above graph shows the variation of electric potential for an electric field directed parallel to the x-axis, and we need to convert it into the graph of electric field strength with x-axis.

We know that,

E = - dv/dx

From the above graph dv/dx is the slope, therefore E = - slope

For electric potential from -3 to -2:

$E = - \dfrac{(10-10)}{(-3)-(-2)} = 0$ N/c

For electric potential from -2 to 0:

$E = - \dfrac{(20-10)}{(0)-(-2)} = -5$ N/c

For electric potential from 0 to +1:

$E = - \dfrac{(20-20)}{(0)-(-1)} = 0$ N/c

For electric potential from +1 to +2:

$E = - \dfrac{(20-10)}{1-2} = +5$ N/c

For electric potential from +2 to +4:

$E = - \dfrac{(20-10)}{4-2} = -5$ N/c

1581018_1751565_ans_b4df4a8096e0433892dfcaabf70f3a9d.png

Electric charge Q is divided into two parts $Q_1$ and $Q_2$ placed at distance r. What would be the condition for the maximum force of repulsion between the charges?

According to question

$Q = Q_1 + Q_2$

$= Q_2 = Q – Q_1$

Coulomb force between two charges,

$F=\dfrac{1}{4\pi \epsilon_0}\times \dfrac{Q_1Q_2}{r^2}$

$\Rightarrow F=\dfrac{1}{4\pi \epsilon_0}\times \dfrac{Q_1(Q-Q_1)}{r^2}$

Where r is the distance between two charges.For F to be maximum.

$\dfrac{dF}{dQ_1}=0$

$F=\dfrac{1}{4\pi \epsilon_0}\times \dfrac{1}{r^2}\dfrac{d}{dQ_1}(QQ_1-{Q_1}^2)=0$

$\dfrac{d}{dQ_1}(QQ_1)-\dfrac{d}{dQ}({Q_1}^2)=0$

$Q-2Q_1=0\,or\,2Q_1=Q$

$Q_1=\dfrac{Q}{2}\,\Rightarrow Q_1=Q_2=\dfrac{Q}{2}$

An electron has an initial velocity of $(12.0\hat{j}+15.0\hat{k}) km/s$ and a constant acceleration of $(2.00\times10^{12}m/s^{2})\hat{i}$ in a region in which uniform electric and magnetic fields are present. If $\vec{B}=(400\mu T)\hat{i}$ , find the electric field $\vec{E}$ .

We apply

$\vec{F}=q(\vec{E}+\vec{v}\times \vec{B})=m_{e}\vec{a}$ to solve for

$\vec{E}$ :

$\vec{E}=\dfrac{m_{e}\vec{a}}{q} + \vec{B}\times\vec{v}$

$=\dfrac{(9.11|times10^{-31}kg)(2.00\times10^{12}m/s^{2})\hat{i}}{-1.60\times10^{-19}C} + (400\mu T)\hat{i} \times [(12.0km/s)\hat{j} + (15.0km/s)\hat{k}]$

$=(-11.4\hat{i}-600\hat{j}+4.80\hat{k})V/m$ .

Analysis Model: Particle in a Field (Electric)
A small object of mass $3.80 \,g$ and charge $-18.0 \mu C$ is suspended motionless above the ground when immersed in a uniform electric field perpendicular to the ground. What are the magnitude and direction of the electric field?

In order for the object to "float" in the electric field, the electric force exerted on the object by the field must be directed upward and have a magnitude equal to the weight of the object. Thus,

$F_e=qE =mg$ , and the magnitude of the electric field must be

$E=\dfrac{mg}{|q|}=\dfrac{(3.80\times 10^{-3} Kg)(9.80 m/s^2)}{1.80\times 10^{-6} C}=\boxed{2.07\times 10^{3} N/C}$
The electric force on a negatively charged object is in the direction opposite to that of the electric field. Since the electric force must be directed.

$\boxed{downward}$

Three equal positive charges $q$ are at the corners of an equilateral triangle of side a as shown in Figure. Assume the three charges together create an electric field. (a) Sketch the field lines in the plane of the charges.(b) Find the location of one point (other than $\infty$ ) where the electric field is zero. What are (c) the magnitude and (d) the direction of the electric field at P due to the two charges at the base?

(a) The electric field has the general appearance shown in figure.
(b) It is zero

$\boxed{\text{at the center}}$ , where (by symmetry) one can see that the three charges individually produce fields that cancel out.
I addition to the center of the triangle, the electric field lines in the second panel of (in figure) indicate three other points near the middle each leg of the triangle where

$E=0$ , but they are more difficult to find mathematically.
(c) You may need to review vector addition in chapter 1. The electric field at point

$P$ can be found by adding the electric field vectors due to each of the two lower point charges:

$\overrightarrow{E}=\overrightarrow{E}_1+\overrightarrow{E}_2$ .
The electric field from a point charge is

$\overrightarrow{E}=k_e\dfrac{q}{r^2}\hat{r}$
As shown in the bottom panel of figure.

$\overrightarrow{E}=k_e\dfrac{q}{a^2}$
to the right and upward at

$60^{o}$ , and

$\overrightarrow{E}_2=k_e\dfrac{q}{a^2}$
to the left and upward at

$60^{o}$ . So,

$\overrightarrow{E}=\overrightarrow{E}_1+\overrightarrow{E}_2=k_e\dfrac{q}{a^2}\left[(\cos 60^{o}\hat{i}+\sin 60^{o}\hat{j})+(-\cos 60^{o}\hat{i}+\sin 60^{o}\hat{j})\right]$

$=k_e\dfrac{q}{a^2}\left[2(\sin 60^{o}\hat{j})\right]=\boxed{1.73 k_e\dfrac{q}{a^2}\hat{j} \,}$
1888735_1859686_ans_43fd96f7793640ccb588cfca09197596.png

A solid insulating sphere of radius $5 cm$ carries electric charge uniformly distributed throughout its volume.Concentric with the sphere is a conducting spherical shell with no net charge as shown in Figure O Q24.The inner radius of the shell is $10 cm$ , and the outer radius is $15 cm$ . No other charges are nearby. Rank the magnitude of the electric field at points $A$ (at radius
$4 cm$ ), $B$ (radius $8 cm$ ), $C$ (radius $12 cm$ ), and $D$ (radius $16 cm$ ) from largest to smallest.Display any cases of equality in your ranking.

The ranking is

$A >B >D >C$ . Let

$q$ represent the charge of the insulating sphere The field at

$A$ is

$(4/5)^{3}q/[4\pi (4cm)^{2}\epsilon_{0}]$ . The field at

$B$ is

$q/[4\pi (8cm)^{2}\epsilon_{0}]$ . The field at

$C$ is zero. The field at

$D$ is

$q/[4\pi (16cm)^{2}\epsilon_{0}]$

A uniformly charged rod of length L and total charge $Q$ lies along the x axis as shown in Figure. (a) Find the components of the electric field at the point $P$ on the y axis a distance $d$ from the origin. (b) What are the approximate values of the field components when $d >> L$ ? Explain why you would expect these results.

(a) The electric field at point

$P$ due to each element of length

$dx$ is

$dE=\dfrac{k_e dq}{x^2+d^2}$ and is directed along the line joining the element to point

$P$ . The charge element

$dq=Q dx/L$ . The

$x$ and

$y$ components are

$E_x=\int dE_x=\int dE\sin \theta$
where

$\sin \theta=\dfrac{x}{\sqrt{d^2+x^2}}$
and

$E_y=\int dE_y=\int dE \cos\theta$ where

$\cos\theta=\dfrac{d}{\sqrt{d^2+x^2}}$
Therefore,

$E_x=-k_e\dfrac{Q}{L}\overset{L}{\underset{0}{\int}}\dfrac{xdx}{(d^2+x^2)^{3/2}}=-k_e\dfrac{Q}{L}\left[\dfrac{-1}{(d^2+x^2)^{1/2}}\right]|^{L}_{0}$

$E_x=-k_e\dfrac{Q}{L}\left[\dfrac{-1}{(d^2+L^2)^{1/2}}-\dfrac{-1}{(d^2+0)^{1/2}}\right]$

$E_x=\boxed{-k_e\dfrac{Q}{L}\left[\dfrac{1}{d}-\dfrac{1}{(d^2+L^2)^{1/2}}\right]}$
and

$E_y=-k_e\dfrac{Qd}{L}\overset{L}{\underset{0}{\int}}\dfrac{dx}{(d^2+x^2)^{3/2}}=k_e\dfrac{Qd}{L}\left[\dfrac{x}{d^2(d^2+x^2)^{1/2}}\right]|^{L}_{0}$

$E_y=k_e\dfrac{Q}{Ld}\left[\dfrac{L}{(d^2+L^2)^{1/2}}-0\right]\rightarrow \, \,E_y=\boxed{k_e\dfrac{Q}{d}\dfrac{1}{(d^2+L^2)^{1/2}}}$
Where

$d>>L$

$E_x=-k_e\dfrac{Q}{L}\left[\dfrac{1}{d}-\dfrac{1}{(d^2+L^2)^{1/2}}\right]\rightarrow -k_e\dfrac{Q}{L}\left[\dfrac{1}{d}-\dfrac{1}{(d^2)^{1/2}}\right]\rightarrow\boxed{E_x\approx 0}$
and

$E_y=k_e\dfrac{Q}{d}\dfrac{1}{(d^2+L^2)^{1/2}}\rightarrow k_e\dfrac{Q}{d}-\dfrac{1}{(d^2)^{1/2}}\rightarrow\boxed{E_y\approx\dfrac{Q}{d^2}}$

$\boxed{\text{which is the field of a point chrage $$Q$$ at a distance $$d$$ along the $$y$$ axis above the charge.}}$
1888712_1859576_ans_dbc8eaa74f814abcb03b5dfb9a1255dd.png

1888712_1859576_ans_dbc8eaa74f814abcb03b5dfb9a1255dd.png

Challenge Problems (89)
A line of charge with uniform density $35.0 \,nC/m$ lies
along the line $y = -15.0 \,cm$ between the points with
coordinates $x = 0$ and $x = 40.0 \,cm$ . Find the electric
field it creates at the origin.

1981954_1861223_ans_aa13359ebd924ebb91bb9815f26262dc.jpg

A solid insulating sphere of radius $5 cm$ carries electric charge uniformly distributed throughout its volume.Concentric with the sphere is a conducting spherical shell with no net charge as shown in Figure O Q24.The inner radius of the shell is $10 cm$ , and the outer radius is $15 cm$ . No other charges are nearby. Similarly rank the electric flux through concentric spherical surfaces through points $A$ , $B$ , $C$ , and $D$ .

The ranking is

$B =D >A >C$ . The flux through the

$4cm$ sphere is

$(4/5)^{3}q/\epsilon_{0}$ . The flux through the

$8cm$ sphere and through the

$16-cm$ sphere is

$q/\epsilon_{0}$ because they enclose the same amount of charge The flux through the

$12-cm$ sphere is

$0$ because the field is zero inside the conductor.

Conductors in Electrostatic Equilibrium
A long, straight metal rod has a radius of $5.00 cm$ and a charge per unit length of $30.0 nC/m$ . Find the electric field (a) $3.00 cm$ , (b) $10.0 cm$ , and (c) $100 cm$ from the axis of the rod, where distances are measured perpendicular to the rods axis.

1974806_1862180_ans_17b5f74b400649cba020a8d227ac2fdb.jpg

A uniform electric field $E = 3 000 \,V/m$ exists within a certain region. What volume of space contains energy equal to $1.00 \times 10^{-7} J$ ? Express your answer in cubic meters and in liters.

1974707_1863452_ans_1a695040ce4b4611a854be8ffe3d46cd.jpg

A current density of $6.00\times 10^{-13} A/m^{ }$ exists in the atmosphere at a location where the electric field is $100 V/m$ . Calculate the electrical conductivity of the Earth’s atmosphere in this region.

1974722_1863495_ans_a056bc9bcfba4319bb446d6a39b8f271.jpeg

Two identical conducting balls A and B have charges $-Q$ and $+3Q$ respectively. they are brought in contact with each other and then separated by a distance d apart. Find the nature of the Coulomb force between them.

$\textbf{Hint: Use Coulomb’s law of electrostatics }$

$\textbf{Step I: Find the charge on each sphere after they are brought in contact with each other }$

$\text{Two identical conducting balls A and B have initial charges -q and +3q respectively.}$

After bringing them in contact both the balls get equal charges as their final potential will be same and due to identical charges final charge will also same

So, Charge on both shell after their contact is

$\frac{-q+3q}{2}$ =+q

$Q_{1}=+q, Q_{2}=+q, r=d$

Step 2: Now applying Coulomb’s law

$F=\dfrac{k\times Q_{1}\times Q_{2}}{r^{2}}$

$F=\dfrac{kq^{2}}{d^{2}}$

$\textbf{Hence, The nature of coulomb force between them is repulsive in nature due to positive charge on both shell}$

Electric Charges And Fields - Class 12 Engineering Physics - Extra Questions

Find the excess (equal in number) of electrons that must be placed on each of two small spheres spaced 3 cm apart with a force of repulsion between the spheres to be 10−19N10^{-19}N

The electric force on a charge q1q_1 due to q2q_2 is (4ˆi−3ˆj)(4\widehat{i} - 3\widehat{j}) N. What is the force on q2q_2 due to q1q_1?

Does the force on a charge due to another charge depend on the charges present nearby?

Is there any lower limit to the electric force between two particles placed at a certain distance?

Find the coulombic force between two α\alpha - particles separated by a distance of 8.2 fermi, in air.

Two charged particles having charge 2×10−8C2\times 10^{-8} C each are joined by an insulating string of length 100 cm. The system is kept on a horizontal table. Find the tension in the string.

Two point charges q1q_1 and q2q_2 are 3 m apart and their combined charge is 20μC20\mu C. If one repels the other with a force of 0.075 N, what are two charges?

Two identical spheres having charges +q1+q_1 and +q2+q_2 are kept at a distance "d". These two are brought in contact with an insulated handle and again separated to the same distance. Find the force F2F_2 between them in terms of initial force F1F_1.

What is the force between two small charged spheres having charges of 2×10−72\times 10^{-7}C and 3×10−73\times 10^{-7}C placed 30 cm apart in air?

The electrostatic force on a small sphere of charge 0.4μC0.4\mu C due to another small sphere of charge −0.8μC-0.8\mu C in air is 0.2 N.(a) What is the distance between the two spheres? (b) What is the force on the second sphere due to the first?

Calculate the height of the potential barrier for a head on collision of two deuterons.

The electrostatic force on a small sphere of charge 0.4 μC0.4\ \mu C due to another small sphere of charge 0.8 μC0.8\ \mu C in air is 0.2 N0.2\ N. (a) What is the distance between the two spheres?(b) What is the force on the second sphere due to the first?

Two point charges of 2×10−7C2\times 10^{-7}C and 1.0×10−71.0\times 10^{-7} are 1.0 cm1.0\ cm apart. What is the magnitude of the field produced by either charge at the site of the other? Use standard value of 1/4πϵ01/4\pi \epsilon_{0}

The electrostatic force on a small sphere of charge 0.4μC0.4 \mu C due to another small sphere of charge −0.8μC-0.8\mu C in the air is 0.2 N0.2\ N.What is the force on the second sphere due to the first?

The electrostatic force on a small sphere of charge 0.4μC0.4\mu C due to another small sphere of charge −0.8μC-0.8\mu C in the air is 0.2 N0.2\ N.What is the distance between the two spheres?

Two equal charges placed in air and separated by a distance of 2 m2\ m repel each other with a force of 10−4kgf10^{-4}kgf. Calculate the magnitude of either of the charges.

The electrostatic force of repulsion between two equal positively charged ions is 3.7×10−9N3.7\times 10^{-9}N, when they are separated by a distance of 5∘A5\overset {\circ}{A}. How many electrons are missing from each ion?

A free pith-ball of 8 g8\ g carries a positive charge of 5×10−8C5\times 10^{-8}C. What must be the nature and magnitude of charge that should be given to a second pith-ball fixed 5 cm5\ cm vertically below the former pith-ball so that the upper pith-ball is stationary?

Two pith-balls of same weight are suspended from the same point by means of silk threads of equal length. On charging the pith-balls equally, they are found to repel each other to a certain distance and remain at equilibrium. Identify the forces responsible for them to remain under equilibrium.

Find the electric force between two protons separated by a distance of 11 fermi (11 fermi=10−15m = {10^{ - 15}}m). The protons in a nucleus remain at a separation of this order.

Calculate the force between two charges each of 20milli20 milli coulomb separated by a distance of 20cm.20 cm. in air.

If the distance between two point charges is increased by 33%, then calculate the % decrease in the force between them.

Two point charges 3C and -6C are 3 m apart. Find the force between them. State the nature of the force.

A charge of 10μc10\mu c and −10μc - 10\mu c are placed 1cm apart. A third charge of is to be placed on the line passing through the two charges such that it is at equilibrium. Find the position of the third point. How does your answer change if the magnitude of the third charge is double?

Calculate the intensity of the electric field due to a helium nucleus at a distance of 1∘A1\overset {\circ}{A} from the nucleus.

Two point charges separated by some distance, repel each other with a force FF, what will be the force if the distance between them is halved?

Two particles with charges q1q_1 and q2q_2 are kept at a certain distance to exert force F on the other. If the distance is reduced to one-fifth, then the force between them is :

under mutual gravitational and electric force then calculate the order of QM\frac{Q}{M} in SI units. The force between two point charges is 100 N in air. Calculate the force if the distance between them is increased by 50%.

Two charge are placed at some separation in vacuum experiences a force FF. If a charge are halfed with separation double and are placed in a medium of dielective constant 44. Find the force experience b the

A charge q=1 μ\mu C is placed at point (1m,2m,4m),. find the electric field at point P(0m,-4m,3m).

Two charges of value 2 μC 2 \ \mu C and −50μC -50 \mu C are placed 80 cm apart. Calculate the distance of the point from the smaller charge where the intensity is zero.

Two equal charges are placed in air are separated by a distance of 3 meter ,repel each other with a force of 0.19f0.19\,f.calculate the magnitude of either of the charge (9.9×10−1C)9.9 \times 10^{-1}C)

The point-like charges carrying charges of +3×10−9C+3\times 10^{-9} C and −5×10−9C-5\times 10^{-9} C are 2 m2 \ m apart. Determine the magnitude of the force between them and state whether it is attractive or repulsive.

The force between two electrons when placed in air is equal to 0.50.5 times the weight of an electron. Find the distance between two electrons (mass of electron =9.1×10−31kg=9.1\times {10}^{-31}kg)

A charge QQ is to be divided on two objects.What should be thevalues of the charges on the two objects so that the force between the objectscan be maximum?

Two point charges, five time as strong as the other repel each other with a force of 1.8×103 1.8 \times 10^3 N. When separated by a distance 1 m, calculate the value of charges?

Three equal charges, 2⋅0×10−6C2\cdot 0\times { 10 }^{ -6 }C each, are held fixed at the three corners of an equilateral triangle, of side 5 cm5\ cm. Find the Coulomb force experienced by one of the charges, due to the other two.

A metal plate of area 0.01 m20.01\ m^{2} carries a charge of 100 μC100\ \mu C. Calculate the outward pull on one side of the plate. [k=1][k=1]

Two point charges of +16μC+16\mu C and −9μC-9\mu C are placed 8 cm8\ cm apart in air. Determine the position of the point at which the resultant field is zero.

Plot a graph showing the variation of coulomb force (F)(F) versus (1r2)\left (\dfrac {1}{r^{2}}\right ) where rr is the distance between the two charges of each pair of charges : (1μC,2μC)(1\mu C, 2\mu C) and (2μC,−3muC)(2\mu C, -3mu C). Interpret the graphs obtained.

A particle of charge 2μC2 \mu C and mass 1.6g1.6 g is moving with a velocity 4ˆims−14 \hat{i} ms^{-1} . At t=0t = 0 the particle enters in a region having an electric field →E\vec{E} (in NC−1)=80ˆi+60ˆjNC^{-1}) = 80 \hat{i} + 60 \hat{j}. Find the velocity of the particle at t=5st = 5 s.

Two point charges of +1μC+1 \mu C and +4μC+4 \mu C are kept 30cm30 cm apart. How far from the +1μC+1 \mu C charge on the line joining the two charge, will the net electric field be zero ?

A pendulum bob of mass 8080mg and carrying a charge of 2×10−82\times 10^{-8}C is at rest in a uniform, horizontal electric field of 20kVm−120 kVm^{-1}. Find the tension in the thread.

A particle of mass m and charge q is thrown at a speed u against a uniform electric field E. How much distance will it travel before coming to momentary rest?

A particle having a charge of 2.0×10−42.0\times 10^{-4}C is placed directly below and at a separation of 1010cm from the bob of a simple pendulum at rest. The mass of the bob is 100100g. What charge should the bob be given so that the string becomes loose?

A particle of mass 11 g and charge 2.5×10−42.5\times 10^{-4}C is released from rest in an electric field of 1.2×104NC−11.2\times 10^{4}NC^{-1}. How long will it take for the particle to travel a distance of 4040cm?

A ball of mass 100100g and having a charge of 4.9×10−54.9\times 10^{-5}C is released from rest in a region where a horizontal electric field of 2.0×104NC−12.0\times 10^4N C^{-1} exists. Find the resulatnt force acting on the ball.

A long cylindrical volume contains a uniformly distributed charge of density ρ\rho. Find the electric field at a point PP inside the cylindrical volume at a distance xx from its axis.

A particle of mass 11 g and charge 2.5×10−42.5\times 10^{-4}C is released from rest in an electric field of 1.2×104NC−11.2\times 10^{4}NC^{-1}. What will be the speed of the particle after travelling the distance of 40cm?

Two particles of charges and masses(+q1,m1) (+q_1, m_1) and (−q2,m2)(-q_2, m_2) are released at different locations in a uniform electric field E in free space. If their separation remains unchanged, find the separation between them.

A ball of mass 100100g and having a charge of 4.9×10−84.9\times 10^{-8}C is released from rest in a region where a horizontal electric field of 2.0×104NC−12.0\times 10^4N C^{-1} exists. Where will the ball be at the end of 2s2s?

A particle of mass 11 g and charge 2.5×10−42.5\times 10^{-4}C is released from rest in an electric field of 1.2×104NC−11.2\times 10^{4}NC^{-1}. How much is the work done by the electric force on the particle as it moves for 40cm?

A grounded metallic ball of radius r is placed on the centre of a uniformly charged thin insulating disc of radiusR(R>>r). R (R >> r). If total charge on the Disc is Q, find force of electrostatic interaction between the ball and the disc.

A ball of mass 100100g and having a charge of 4.9×10−54.9\times 10^{-5}C is released from rest in a region where a horizontal electric field of 2.0×104NC−12.0\times 10^4N C^{-1} exists. What will be the path of the ball?

Three particles, each with positive charge QQ, form an equilateral triangle, with each side of length dd. What is the magnitude of the electric field produced by the particles at the midpoint of any side?

What is the acceleration of an electron in an uniform electric field of magnitude 1.40×106 N/C1.40\times 10^6\ N/C?

Two tiny spherical water drops with identical charges of −1.00×10−16C-1.00\times 10^{-16}C, have a center-to-center separation of 1.00 cm1.00\ cm.What is the magnitude of the electrostatic force acting between them?

What is the magnitude of the electrostatic force between a singly charged sodium ion (Na+Na^{+}, of charge +e+e) and an adjacent singly charged chlorine ion (Cl−Cl^{-}, of charge −e-e) in a salt crystal if their separation is 2.82×10−10m2.82\times 10^{-10}m?

A charge of 6.0μC6.0 \mu C is to be split into two parts that are then separated by 3.0 mm.3.0\ mm. What is the maximum possible magnitude of the electrostatic force between those two parts?

The charge and coordinates of two charged particles held fixed in an yy plane are q1=+3.0 μC,x1=3.5 cm,y1=0.50 cmq_1=+3.0\ \mu C, x_1=3.5\ cm, y_1=0.50\ cm, and q2=−4.0μC,x2=−2.0 cm,y2=1.5 cmq_2 =-4.0\mu C, x_2 =-2.0\ cm, y_2 =1.5\ cm. Find magnitude?

What would be the magnitude of the electrostatic force between two 1.00 C1.00\ C point charges separated by a distance of 1.00 km1.00\ km if such point charges existed (they do not) and this configuration could be set up?

What would be the magnitude of the electrostatic force between two 1.00 C1.00\ C point charges separated by a distance of 1.00 m1.00\ m

A particle of charge +3.00×10−6C+3.00\times 10^{-6}C is 12.0 cm12.0\ cm distant from a second particle of charge −1.50×10−6C-1.50\times 10^{-6}C. Calculate the magnitude of the electrostatic force between the particles.

A neutron consists of ine "up" quark of charge 2e/32e/3 and two "down" quarks each having charge −e/3.-e/3. If we assume that the down quarks are 26×10−1526 \times 10^{-15}\ m$$ apart inside the neutron, what is the magnitude of the electrostatic force between them?

In Fig, the three particles are fixed in place and have charges q1=q2=+eq_{1}=q_{2}=+e and q3=+2eq_{3}=+2e. Distance a=6.00μma=6.00 \mu m. What are themagnitude

Of the charge QQ on a tiny sphere, a fraction α\alpha is to be transferred to a second, nearby sphere. The spheres can be treated as particle. What value of α\alpha maximizes the magnitude FF of the electrostatic force between the two spheres? What are the

The force between two point charges kept at a distance rr apart in air is FF. It the same charges are kept in water at the same distance, how does the force between them change?

A steady current is flowing in a cylindrical conductor.Does electric field exist within the conductor?

Charge is uniformly distributed around a ring of radius R=2.40 cmR=2.40\ cm, and the resulting field magnitude EE is measure along the ring's central axis (perpendicular to the plane of the ring). At what distance from the ring's centre is EE maximum?

Figure shows three circular centred on the origin of a coordinate system. On each arc, the uniformly distributed charge is given in terms of Q=2.00 μCQ=2.00\ \mu C. The radii are given in term of R=10.0 cmR=10.0\ cm.What are the magnitude?

Is the force acting between two point charges q1q1 and q2q2 kept at some distance apart in air attractive or repulsive when (i) q1q2>0q1q2 > 0 (ii) q1q2<0q1q2 < 0?

A 10.0 g10.0\ g block with a charge of +8.00×10−5C+8.00 \times 10^{-5}C is placed in an electric field →E=(3000ˆi−600ˆj) N/C\vec E=(3000 \hat i-600 \hat j)\ N/C. If the block is released from rest at the origin at time t=0t=0, what is its yy coordinate at t=3.00 st=3.00\ s?

What is the nature of electrostatic force between two point electric charges q1q_{1} and q2q_{2} if(a) q1+q2>0?q_{1} + q_{2} > 0?(b) q1+q2<0?q_{1} + q_{2} < 0?

A 10.0 g10.0\ g block with a charge of +8.00×10−5C+8.00 \times 10^{-5}C is placed in an electric field →E=(3000ˆi−600ˆj) N/C\vec E=(3000 \hat i-600 \hat j)\ N/C. If the block is released from rest at the origin at time t=0t=0, what is its xx coordinate at t=3.00 st=3.00\ s?

In Fig the three particles are fixed in place and have charges q1=q2=+eq_{1}=q_{2}=+e and q3=+2eq_{3}=+2e. Distance a=6.00μma=6.00 \mu m. What are thedirection of the net electric field at point PP due to the particles?

A 10.0 g10.0\ g block with a charge of +8.00×10−5C+8.00 \times 10^{-5}C is placed in an electric field →E=(3000ˆi−600ˆj) N/C\vec E=(3000 \hat i-600 \hat j)\ N/C. What is the magnitude.

In figure, a nonconducting rod of length L=8.15 cmL=8.15\ cm has a charge −1=−4.23 fC-1=-4.23\ fC uniformly distributed along its length.What are the magnitude ?

Five charges, qq each is placed at the corners of a regular pentagon of side aa as shown in figure. What will be the electric field at OO, the centre of the regular pentagon?

Can electric field at a point be zero, while electric potential is not zero

Find the excess (equal in number) of electrons that must be placed on each of two small spheres spaced 3 cm apart with a force of repulsion between the spheres to be $10^{-19}N$

The electric force on a charge $q_1$ due to $q_2$ is $(4\widehat{i} - 3\widehat{j})$ N. What is the force on $q_2$ due to $q_1$ ?

Find the coulombic force between two $\alpha$ - particles separated by a distance of 8.2 fermi, in air.

Two charged particles having charge $2\times 10^{-8} C$ each are joined by an insulating string of length 100 cm. The system is kept on a horizontal table. Find the tension in the string.

Two point charges $q_1$ and $q_2$ are 3 m apart and their combined charge is $20\mu C$ . If one repels the other with a force of 0.075 N, what are two charges?

Two identical spheres having charges $+q_1$ and $+q_2$ are kept at a distance "d". These two are brought in contact with an insulated handle and again separated to the same distance. Find the force $F_2$ between them in terms of initial force $F_1$ .

What is the force between two small charged spheres having charges of $2\times 10^{-7}$ C and $3\times 10^{-7}$ C placed 30 cm apart in air?

The electrostatic force on a small sphere of charge $0.4\mu C$ due to another small sphere of charge $-0.8\mu C$ in air is 0.2 N.
(a) What is the distance between the two spheres?
(b) What is the force on the second sphere due to the first?

The electrostatic force on a small sphere of charge $0.4\ \mu C$ due to another small sphere of charge $0.8\ \mu C$ in air is $0.2\ N$ .
(a) What is the distance between the two spheres?
(b) What is the force on the second sphere due to the first?

Two point charges of $2\times 10^{-7}C$ and $1.0\times 10^{-7}$ are $1.0\ cm$ apart. What is the magnitude of the field produced by either charge at the site of the other? Use standard value of $1/4\pi \epsilon_{0}$

The electrostatic force on a small sphere of charge $0.4 \mu C$ due to another small sphere of charge $-0.8\mu C$ in the air is $0.2\ N$ .
What is the force on the second sphere due to the first?

The electrostatic force on a small sphere of charge $0.4\mu C$ due to another small sphere of charge $-0.8\mu C$ in the air is $0.2\ N$ .
What is the distance between the two spheres?

Two equal charges placed in air and separated by a distance of $2\ m$ repel each other with a force of $10^{-4}kgf$ . Calculate the magnitude of either of the charges.

The electrostatic force of repulsion between two equal positively charged ions is $3.7\times 10^{-9}N$ , when they are separated by a distance of $5\overset {\circ}{A}$ . How many electrons are missing from each ion?

A free pith-ball of $8\ g$ carries a positive charge of $5\times 10^{-8}C$ . What must be the nature and magnitude of charge that should be given to a second pith-ball fixed $5\ cm$ vertically below the former pith-ball so that the upper pith-ball is stationary?

Find the electric force between two protons separated by a distance of $1$ fermi ( $1$ fermi $= {10^{ - 15}}m$ ). The protons in a nucleus remain at a separation of this order.

Calculate the force between two charges each of $20 milli$ coulomb separated by a distance of $20 cm.$ in air.

If the distance between two point charges is increased by $3%$ , then calculate the $%$ decrease in the force between them.

A charge of $10\mu c$ and $- 10\mu c$ are placed 1cm apart. A third charge of is to be placed on the line passing through the two charges such that it is at equilibrium. Find the position of the third point. How does your answer change if the magnitude of the third charge is double?

Calculate the intensity of the electric field due to a helium nucleus at a distance of $1\overset {\circ}{A}$ from the nucleus.

Two point charges separated by some distance, repel each other with a force $F$ , what will be the force if the distance between them is halved?

Two particles with charges $q_1$ and $q_2$ are kept at a certain distance to exert force F on the other. If the distance is reduced to one-fifth, then the force between them is :

under mutual gravitational and electric force then calculate the order of $\frac{Q}{M}$ in SI units.
The force between two point charges is 100 N in air. Calculate the force if the distance between them is increased by 50%.

Two charge are placed at some separation in vacuum experiences a force $F$ . If a charge are halfed with separation double and are placed in a medium of dielective constant $4$ . Find the force experience b the

A charge q=1 $\mu$ C is placed at point (1m,2m,4m),. find the electric field at point P(0m,-4m,3m).

Two charges of value $2 \ \mu C$ and $-50 \mu C$ are placed 80 cm apart. Calculate the distance of the point from the smaller charge where the intensity is zero.

Two equal charges are placed in air are separated by a distance of 3 meter ,repel each other with a force of $0.19\,f$ .calculate the magnitude of either of the charge ( $9.9 \times 10^{-1}C)$

The point-like charges carrying charges of $+3\times 10^{-9} C$ and $-5\times 10^{-9} C$ are $2 \ m$ apart. Determine the magnitude of the force between them and state whether it is attractive or repulsive.

The force between two electrons when placed in air is equal to $0.5$ times the weight of an electron. Find the distance between two electrons (mass of electron $=9.1\times {10}^{-31}kg$ )

A charge $Q$ is to be divided on two objects.What should be the
values of the charges on the two objects so that the force between the objects
can be maximum?

Two point charges, five time as strong as the other repel each other with a force of $1.8 \times 10^3$ N. When separated by a distance 1 m, calculate the value of charges?

Three equal charges, $2\cdot 0\times { 10 }^{ -6 }C$ each, are held fixed at the three corners of an equilateral triangle, of side $5\ cm$ . Find the Coulomb force experienced by one of the charges, due to the other two.

A metal plate of area $0.01\ m^{2}$ carries a charge of $100\ \mu C$ . Calculate the outward pull on one side of the plate. $[k=1]$

Two point charges of $+16\mu C$ and $-9\mu C$ are placed $8\ cm$ apart in air. Determine the position of the point at which the resultant field is zero.

Plot a graph showing the variation of coulomb force $(F)$ versus $\left (\dfrac {1}{r^{2}}\right )$ where $r$ is the distance between the two charges of each pair of charges : $(1\mu C, 2\mu C)$ and $(2\mu C, -3mu C)$ . Interpret the graphs obtained.

A particle of charge $2 \mu C$ and mass $1.6 g$ is moving with a velocity $4 \hat{i} ms^{-1}$ . At $t = 0$ the particle enters in a region having an electric field $\vec{E}$ (in $NC^{-1}) = 80 \hat{i} + 60 \hat{j}$ . Find the velocity of the particle at $t = 5 s$ .

Two point charges of $+1 \mu C$ and $+4 \mu C$ are kept $30 cm$ apart. How far from the $+1 \mu C$ charge on the line joining the two charge, will the net electric field be zero ?

A pendulum bob of mass $80$ mg and carrying a charge of $2\times 10^{-8}$ C is at rest in a uniform, horizontal electric field of $20 kVm^{-1}$ . Find the tension in the thread.

A particle having a charge of $2.0\times 10^{-4}$ C is placed directly below and at a separation of $10$ cm from the bob of a simple pendulum at rest. The mass of the bob is $100$ g. What charge should the bob be given so that the string becomes loose?

A particle of mass $1$ g and charge $2.5\times 10^{-4}$ C is released from rest in an electric field of $1.2\times 10^{4}NC^{-1}$ . How long will it take for the particle to travel a distance of $40$ cm?

A ball of mass $100$ g and having a charge of $4.9\times 10^{-5}$ C is released from rest in a region where a horizontal electric field of $2.0\times 10^4N C^{-1}$ exists. Find the resulatnt force acting on the ball.

A long cylindrical volume contains a uniformly distributed charge of density $\rho$ . Find the electric field at a point $P$ inside the cylindrical volume at a distance $x$ from its axis.

A particle of mass $1$ g and charge $2.5\times 10^{-4}$ C is released from rest in an electric field of $1.2\times 10^{4}NC^{-1}$ . What will be the speed of the particle after travelling the distance of 40cm?

Two particles of charges and masses $(+q_1, m_1)$ and $(-q_2, m_2)$ are released at different locations in a uniform electric field E in free space. If their separation remains unchanged, find the separation between them.

A ball of mass $100$ g and having a charge of $4.9\times 10^{-8}$ C is released from rest in a region where a horizontal electric field of $2.0\times 10^4N C^{-1}$ exists. Where will the ball be at the end of $2s$ ?

A particle of mass $1$ g and charge $2.5\times 10^{-4}$ C is released from rest in an electric field of $1.2\times 10^{4}NC^{-1}$ . How much is the work done by the electric force on the particle as it moves for 40cm?

A grounded metallic ball of radius r is placed on the centre of a uniformly charged thin insulating disc of radius $R (R >> r).$ If total charge on the Disc is Q, find force of electrostatic interaction between the ball and the disc.

A ball of mass $100$ g and having a charge of $4.9\times 10^{-5}$ C is released from rest in a region where a horizontal electric field of $2.0\times 10^4N C^{-1}$ exists. What will be the path of the ball?

Three particles, each with positive charge $Q$ , form an equilateral triangle, with each side of length $d$ . What is the magnitude of the electric field produced by the particles at the midpoint of any side?

What is the acceleration of an electron in an uniform electric field of magnitude $1.40\times 10^6\ N/C$ ?

Two tiny spherical water drops with identical charges of $-1.00\times 10^{-16}C$ , have a center-to-center separation of $1.00\ cm$ .
What is the magnitude of the electrostatic force acting between them?

What is the magnitude of the electrostatic force between a singly charged sodium ion ( $Na^{+}$ , of charge $+e$ ) and an adjacent singly charged chlorine ion ( $Cl^{-}$ , of charge $-e$ ) in a salt crystal if their separation is $2.82\times 10^{-10}m$ ?

A charge of $6.0 \mu C$ is to be split into two parts that are then separated by $3.0\ mm.$ What is the maximum possible magnitude of the electrostatic force between those two parts?

The charge and coordinates of two charged particles held fixed in an $y$ plane are $q_1=+3.0\ \mu C, x_1=3.5\ cm, y_1=0.50\ cm$ , and $q_2 =-4.0\mu C, x_2 =-2.0\ cm, y_2 =1.5\ cm$ . Find
magnitude?

What would be the magnitude of the electrostatic force between two $1.00\ C$ point charges separated by a distance of $1.00\ km$ if such point charges existed (they do not) and this configuration could be set up?

What would be the magnitude of the electrostatic force between two $1.00\ C$ point charges separated by a distance of $1.00\ m$

A particle of charge $+3.00\times 10^{-6}C$ is $12.0\ cm$ distant from a second particle of charge $-1.50\times 10^{-6}C$ . Calculate the magnitude of the electrostatic force between the particles.

A neutron consists of ine "up" quark of charge $2e/3$ and two "down" quarks each having charge $-e/3.$ If we assume that the down quarks are $26 \times 10^{-15}$ \ m$$ apart inside the neutron, what is the magnitude of the electrostatic force between them?

In Fig, the three particles are fixed in place and have charges $q_{1}=q_{2}=+e$ and $q_{3}=+2e$ . Distance $a=6.00 \mu m$ . What are the
magnitude

Of the charge $Q$ on a tiny sphere, a fraction $\alpha$ is to be transferred to a second, nearby sphere. The spheres can be treated as particle. What value of $\alpha$ maximizes the magnitude $F$ of the electrostatic force between the two spheres? What are the

The force between two point charges kept at a distance $r$ apart in air is $F$ . It the same charges are kept in water at the same distance, how does the force between them change?

Charge is uniformly distributed around a ring of radius $R=2.40\ cm$ , and the resulting field magnitude $E$ is measure along the ring's central axis (perpendicular to the plane of the ring). At what distance from the ring's centre is $E$ maximum?

Figure shows three circular centred on the origin of a coordinate system. On each arc, the uniformly distributed charge is given in terms of $Q=2.00\ \mu C$ . The radii are given in term of $R=10.0\ cm$ .
What are the magnitude?

Is the force acting between two point charges $q1$ and $q2$ kept at some distance apart in air attractive or repulsive when (i) $q1q2 > 0$ (ii) $q1q2 < 0$ ?

A $10.0\ g$ block with a charge of $+8.00 \times 10^{-5}C$ is placed in an electric field $\vec E=(3000 \hat i-600 \hat j)\ N/C$ . If the block is released from rest at the origin at time $t=0$ , what is its $y$ coordinate at $t=3.00\ s$ ?

What is the nature of electrostatic force between two point electric charges $q_{1}$ and $q_{2}$ if
(a) $q_{1} + q_{2} > 0?$
(b) $q_{1} + q_{2} < 0?$

A $10.0\ g$ block with a charge of $+8.00 \times 10^{-5}C$ is placed in an electric field $\vec E=(3000 \hat i-600 \hat j)\ N/C$ . If the block is released from rest at the origin at time $t=0$ , what is its $x$ coordinate at $t=3.00\ s$ ?

In Fig the three particles are fixed in place and have charges $q_{1}=q_{2}=+e$ and $q_{3}=+2e$ . Distance $a=6.00 \mu m$ . What are the
direction of the net electric field at point $P$ due to the particles?

A $10.0\ g$ block with a charge of $+8.00 \times 10^{-5}C$ is placed in an electric field $\vec E=(3000 \hat i-600 \hat j)\ N/C$ . What is the magnitude.

In figure, a nonconducting rod of length $L=8.15\ cm$ has a charge $-1=-4.23\ fC$ uniformly distributed along its length.
What are the magnitude ?

Five charges, $q$ each is placed at the corners of a regular pentagon of side $a$ as shown in figure. What will be the electric field at $O$ , the centre of the regular pentagon?

Five charges, $q$ each is placed at the corners of n-sided regular polygon of side $a$ . What will be the electric field at $O$ , the centre of the n-sided regular polygon? (See figure for analogy)

Five charges, $q$ each is placed at the corners of a regular pentagon of side $a$ as shown in figure. What will be the electric field at $O$ if the charge from one of the corners (say $A$ ) is removed?

Five charges, $q$ each is placed at the corners of an n-sided regular polygon of side $a$ (see figure for analogy). What will be the electric field at $O$ if the charge $q$ at A is replaced by $-q$ ?

For two point charges and $q_1$ and $q_2$ , $q_1q_2 > 0$ . What is the nature of electrostatic force between two charges?

For two point charges $q_1$ and $q_2$ , $q_1q_2<0$ . What is the nature of electrostatic force between two charges?

The electric potential at a point $(x, y)$ in an electric field is given by $V=6 x y+y^{2}-x^{2}$ Calculate the electric field at that point.

Two point charges $q_A=3\ \mu C$ and $q_B=3\ \mu C$ are located $20\ cm$ apart in vacuum.
If a negative test charge of magnitude $1.5\times 10^{-9}C$ is placed at this point, what is the force experienced by the test charge?

What is the force between two small charged spheres having charges of $2\times 10^{-7}$ and $3\times 10^{-7} C$ placed $30\ cm$ apart in air?

A non-polar molecule with polarizability $\beta$ is located at a great distance $l$ from a polar molecule with electric moment $p$ . Find the magnitude of the interaction force between the molecules if the vector $p$ is oriented along a straight line passing through both molecules.