MCQ Questions for Class 12 Medical Physics Electric Charges And Fields Quiz 8

Two charges

$+20\mu C$ and

$-20\mu C$ are placed

$10mm$ apart. The electric field at point

$P$ , on the axis of the dipole

$10cm$ away from its centre

$O$ on the side of the positive charge is

0%
$8.6\times { 10 }^{ 9 }N\quad { C }^{ -1 }$
0%
$4.1\times { 10 }^{ 6 }N\quad { C }^{ -1 }$
0%
$3.6\times { 10 }^{ 6 }N\quad { C }^{ -1 }$
0%
$4.6\times { 10 }^{ 5 }N\quad { C }^{ -1 }$

Explanation

Here

$\quad q=\pm 20\mu C=\pm 20\times { 10 }^{ -6 }C,2a=10mm=10\times { 10 }^{ -3 }m,r=OP=10cm=10\times { 10 }^{ -2 }m$

$\left| \vec { p } \right| =q\times 2a=20\times { 10 }^{ -6 }\times 10\times { 10 }^{ -3 }m=2\times { 10 }^{ -7 }m\quad$
The electric field along

$BP,\vec { E } =\cfrac { 2\vec { p } r }{ 4\pi { \varepsilon }_{ 0 }{ \left( { r }^{ 2 }-{ a }^{ 2 } \right) }^{ 2 } }$
As

$a<<r$

$\vec { E } =\cfrac { 2\left| \vec { p } \right| }{ 4\pi { \varepsilon }_{ 0 }{ r }^{ 3 } } =\cfrac { 2\times 2\times { 10 }^{ -7 }\times 9\times { 10 }^{ 9 } }{ { \left( 10\times { 10 }^{ -2 } \right) }^{ 3 } } =3.6\times { 10 }^{ 6 }N{ C }^{ -1 }\quad$
1028045_941055_ans_c0bc9da5053844558ead85fbfcd25c1a.PNG

1028045_941055_ans_c0bc9da5053844558ead85fbfcd25c1a.PNG

Which of the following statements about dipole moment is not true ?

0%
The dimension of dipole moment is $\left[ LTA \right]$
0%
The unit of dipole moment is $C$ $m$
0%
Dipole moment is vector quantity and directed from negative to positive charge
0%
Dipole moment is a scalar quantity and has magnitude charge equal to the potential of separation between charge

Explanation

Dipole moment is a vector quantity and has magnitude of

$2qa$ and it is in the direction of the dipole axis from

$-q$ to

$q$

The electric field intensity at a point

$P$ due to point charge

$q$ kept at point

$Q$ is

$24N{ C }^{ -1 }$ and the electric potential at point is

$12J{ C }^{ -1 }$ . The order of magnitude of charge

$q$ is

0%
${ 10 }^{ -6 }C$
0%
${ 10 }^{ -7 }C$
0%
${ 10 }^{ -10 }C$
0%
${ 10 }^{ -9 }C$

Explanation

Electric field of a point charge,

$E=\cfrac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \cfrac { q }{ { r }^{ 2 } } =24N{ C }^{ -1 }$
Electric potential of a point charge

$V=\cfrac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \cfrac { q }{ { r }^{ } } \quad =12J{ C }^{ -1 }\quad$
The distance

$PQ$ is

$r=\cfrac { V }{ E } =\cfrac { 12 }{ 24 } =0.5m$

$\therefore$ Magnitude of charge

$q'=4\pi { \varepsilon }_{ 0 }{ V }_{ r }=\cfrac { 1 }{ 9\times { 10 }^{ 9 } } \times 12\times 0.5=0.667\times { 10 }^{ -9 }C={ 10 }^{ -9 }C\quad$

Choose the correct answer from the alternatives given.
If

$\varepsilon_0$ and

$\mu_0$ are respectively the electric permittivity and the magnetic permeability of free space and

$\varepsilon$ and

$\mu$ the corresponding quantities in a medium, the refractive index of the medium is

0%
$\sqrt{\dfrac{\mu\varepsilon}{\mu_0 \varepsilon_0}}$
0%
$\dfrac{\mu\varepsilon}{\mu_0 \varepsilon_0}$
0%
$\sqrt{\dfrac{\mu_0\varepsilon_0}{\mu\varepsilon}}$
0%
$\dfrac{\mu\mu_0}{\varepsilon\varepsilon_0}$

Explanation

So, Refractive index of a medium is given by

$n=\dfrac{c}{v}$ .....................(1)

$c$ is the velocity of light in vacuum.

$v$ is the velocity of light in the medium.

Also, speed is given by

$\dfrac{1}{\sqrt{\mu \epsilon}}$

$c$

$=$

$\dfrac{1}{\sqrt{\mu_0 \epsilon_0}}$ ..........(2)

$v$

$=$

$\dfrac{1}{\sqrt{\mu \epsilon}}$ .................(3)

From eq (1), (2) and (3):

$n=\sqrt{\dfrac{\mu \epsilon}{\mu_0 \epsilon_0}}$

The electric field and the potential of an electric dipole vary with distance

$r$ as

0%
$\cfrac{1}{r}$ and $\cfrac { 1 }{ { r }^{ 2 } }$
0%
$\cfrac { 1 }{ { r }^{ 2 } }$ and $\cfrac{1}{r}$
0%
$\cfrac { 1 }{ { r }^{ 2 } }$ and $\cfrac { 1 }{ { r }^{ 3 } }$
0%
$\cfrac { 1 }{ { r }^{ 3 } }$ and $\cfrac { 1 }{ { r }^{ 2 } }$

Explanation

Electric field and electric potential at a general point at a distance r from the centre of the dipole is

$E_{g}=\dfrac{1}{4\pi \epsilon_{o}}.\dfrac{p}{r^3}\sqrt{3\cos^2 \theta+1}$

And,

$V_{g}=\dfrac{1}{4\pi \epsilon_{o}}.\dfrac{p\cos \theta}{r^2}$

Hence, D is the correct option.

Figure shows electric field lines in which an electric dipole

$\vec { P }$ is placed as shown.
Which of the following statements is correct?

0%
The dipole will not experience any force
0%
The dipole will experience a force towards right
0%
The dipole will experience a force towards left
0%
The dipole will experience a force upwards

Explanation

The spacing between electric lines of force increases from left to right. It means that the electric field intensity

$E$ on left is greater than the electric field intensity

$E$ on right.

So, the force on

$+q$ charge of the dipole is smaller and to the right. Force on

$-q$ charge of the dipole is bigger and to the left due to variable electric field. Hence the dipole will experience force towards the left.

Which of the following is not true?

0%
For a point charge, electrostatic potential varies as $\dfrac { 1 }{ r }$ .
0%
For a dipole, the potential depends on the magnitude of position vector and dipole moment vector.
0%
The electric dipole potential varies as $\dfrac { 1 }{ r }$ at large distance.
0%
For a point charge, the electrostatic field varies as $\dfrac { 1 }{{ r }^{ 2 }}$ .

Explanation

The electric potential due to point charge is given by:

$V=\dfrac{1}{4\pi\varepsilon_0}\dfrac{q}{r}$

Therefore, statement

$A$ is true.

The electrostatic field due to a point charge is given by:

$E=\dfrac{1}{4\pi\varepsilon_0}\dfrac{q}{r^2}$

So, statement

$D$ is true.

The potential due to a dipole is given by:

$V_d=\dfrac{1}{4\pi\varepsilon_0}\dfrac{p.\hat{r}}{r^3}$

The above expression shows that the statement

$B$ is correct and statement

$C$ is incorrect.

When air is replaced by a dielectric medium of constant K, the maximum force of attraction between two charges separated by a distance:

0%
increases K times
0%
remains unchanged
0%
decreases K times
0%
increases ${ K }^{ -1 }$ times

Explanation

Hint:-Use the formula of Coulamb's Force in a medium

Explanation:-

$\bullet$ Force of attraction between two charges in air is given by ${{F}_{a}}=\dfrac{q_1q_2}{4\pi {{\varepsilon }_{0}}{{r}^{2}}}$

$\bullet$ When the air is replaced by dielectric constant K, the force of attraction between two charges becomes ${{F}_{m}}=\dfrac{q_1q_2}{4\pi K{{\varepsilon }_{0}}}=\dfrac{{{F}_{a}}}{K}$

Thus the force is decreasing by K times

$\textbf{Hence, the correct option is (C)}$

Choose the correct statement

0%
Polar molecules have permanent electric dipole moment
0%
${CO}_{2}$ molecule is a polar molecule
0%
${H}_{2}O$ is a non-polar molecule
0%
The dipole field at large distances falls of as $\cfrac { 1 }{ { r }^{ 2 } }$

Explanation

The polar molecules do not have a permanent electric dipole moment and this dipole moment does not change because of the external factors.

The

$CO_2$ is a non-polar molecule with no net dipole moment whereas

$H_2O$ is a polar molecule.

The field due to a dipole varies as inversely proportional to the distance of the point from the axis of dipole.

$E\propto\dfrac1r$ .

Therefore, statement

$(A)$ is correct.

Which among the following is an example of polar molecule?

0%
${ O }_{ 2 }$
0%
${ H }_{ 2 }$
0%
${ N }_{ 2 }$
0%
HCl

Explanation

In polar molecule the centres of positive and negative charges are separated even when there is no external field. Such molecule have a permanent dipole moment. Ionic molecule like

$HCl$ is an example of polar molecule.

Two large thin metal plates are parallel and close to each other. On their inner faces, the plates have surface charge densities of opposite signs and magnitude

$27\times { 10 }^{ -22 }C\quad { m }^{ -2 }$ . The electric field

$\vec { E }$ in region II in between the plates is

0%
$4.25\times { 10 }^{ -8 }N\quad { C }^{ -1 }\quad$
0%
$6.28\times { 10 }^{ -10 }N\quad { C }^{ -1 }$
0%
$3.05\times { 10 }^{ -10 }N\quad { C }^{ -1 }$
0%
$5.03\times { 10 }^{ -10 }N\quad { C }^{ -1 }$

Explanation

The value of

$\vec { E }$ in the region II, in between the plates

$=\cfrac { \sigma }{ { \varepsilon }_{ 0 } } =\cfrac { 27\times { 10 }^{ -22 } }{ 8.854\times { 10 }^{ -12 } } =3.05\times { 10 }^{ -10 }N{ C }^{ -1 }$

Two conducting and concentric thin spherical shells or radii

$a$ and

$b,\,\left( {b > a} \right)$ have charges

${q_1}$ and

${q_2}$ respectively. Now if the inner shell is earthed then the final charge on this shell will be

0%
$\dfrac{{{q_2}{a^2}}}{{{b^2}}}$
0%
$\dfrac{{ - {q_2}a}}{b}$
0%
$\dfrac{{\left( {{q_1} + {q_2}} \right)}}{2}$
0%
$- \dfrac{{{q_2}b}}{a}$

Explanation

$\therefore$ The inner spherical conductor is grounded,

$\Rightarrow$ potential,

$V=0$

Also, Potential at inner shell,

$V=\dfrac { K{ Q }_{ 1 } }{ a } +\dfrac { { KQ }_{ 2 } }{ b }$

$\Rightarrow \quad O=\dfrac { K{ Q }_{ 1 } }{ a } +\dfrac { { KQ }_{ 2 } }{ b }$

${ Q }_{ 1 }=\dfrac { -{ Q }_{ 2 }a }{ b }$

$\therefore$ Charge on inner shell

${ Q }_{ 1 }=\dfrac { -{ Q }_{ 2 }a }{ b }$

Option (B) is the correct option.

A uniform line charge with linear density

$\lambda$ lies along the y-axis. What flux crosses a spherical surface centred at he origin with

$r\ =\ R$

0%
$\large{\frac{2R\lambda}{\epsilon_0}}$
0%
$\large{\frac{R\lambda}{\epsilon_0}}$
0%
$\large{\frac{\lambda}{\epsilon_0}}$
0%
none of the above

Explanation

Line charge density

$=\lambda$

radius,

$r=R$

Now,

$\lambda dl=d\theta$

for total charge enclosed by radius R,

$\int _{ O }^{ R }{ Q } =\int _{ O }^{ R }{ \lambda dl }$

$\boxed { { Q }_{ inc }=\lambda R }$

Now,

$\phi =\dfrac { { Q }_{ inc } }{ { \varepsilon }_{ 0 } }$ {from Gauss law}

$=\frac { \lambda R }{ { \varepsilon }_{ 0 } }$ (Option B is the correct option)

The electric field of a radio wave is given by

$\vec{E} = E sin (Kz - \omega t) (\hat{i} + \hat{j})$ . Give a unit vector
in the direction of the magnetic field at a place and time where sin

$(kz - \omega t)$ is positive.

0%
$(\hat{i}^ + \hat{k}) / \sqrt{2}$
0%
$(-\hat{i} - \hat{k} / \sqrt{2}$
0%
$-\hat{i} + \hat{j} / \sqrt{2}$
0%
$\hat{i} - \hat{j} / \sqrt{2}$

Explanation

$\vec{E} = E\sin(Kz-\omega t)(\hat{i} + \hat{j})$

$\hat{E} = \dfrac{\hat{i} + \hat{j}}{\sqrt{1^2 + 1^2}} \\$

$\hat{E} = \dfrac{\hat{i} + \hat{j}}{\sqrt{2}}$

unit vector in the direction of the magnetic field where

$\sin {(kz - \omega t)}$ is positive will be in same plane and perpendicular to

$\hat{E}$

unit vector

$= \dfrac{1}{\sqrt{x_1^2 + y_1^2}} (x_1\hat{i} + y_1\hat{j})$

Both vector are perpendicular to each other hence dot product will be zero.

$\dfrac{1}{\sqrt2} * \dfrac{x_1}{\sqrt{x_1^2 + y_1^2}} + \dfrac{1}{\sqrt2} * \dfrac{y_1}{\sqrt{x_1^2 + y_1^2}} = 0$

$x_1 = - y_1$

$y_1 = - x_1$

unit vector

$= \dfrac{1}{\sqrt{x_1^2 + {-x_1}^2}} (x_1\hat{i} - x_1\hat{j}) or \dfrac{1}{\sqrt{y_1^2 + {-y_1}^2}} (-y_1\hat{i} + y_1\hat{j}) \\ = \dfrac{1}{\sqrt{2x_1^2}} (x_1(\hat{i} - \hat{j})) or = \dfrac{1}{\sqrt{2y_1^2}} (y_1(\hat{-i} + \hat{j})) \\ = \dfrac{1}{\sqrt{2}} (\hat{i} - \hat{j}) or = \dfrac{1}{\sqrt{2}} (-\hat{i} + \hat{j})$

But for

$\dfrac{\hat{i} - \hat{j}}{\sqrt2} \, , \, \sin = -ve$ so this will not possible

Hence

$\dfrac{-\hat{i} + \hat{j}}{\sqrt2}$ is correct answer.

Two concentric spherical shells of radii R and

$2$ R have charges

$+$ Q and

$-$ Q. Which of the following may represent correct variation of potential with distance r from origin?

Explanation

The potential inside the spherical shell is constant

For point A

$P_A=\cfrac{K Q}{R}$ and

For point B

$P_B=\cfrac{K Q}{2R} +\cfrac{K Q}{R}$

So the correct answer is 2.

1137612_996164_ans_5c10ec82a1224c499d7ef175c59b61b8.png

Two equal point charges are fixed at

$x=-a$ and

$x=+a$ on the

$x-$ axis. Another point charge

$Q$ is placed at the origin. The change in the electrical potential energy of

$Q$ , when it is displaced by a small distance

$x$ along the

$x-$ axis, is approximately proportional to:

0%
$x$
0%
$x^{2}$
0%
$x^{3}$
0%
$\dfrac{1}{x}$

Explanation

Potential energy $=\cfrac{1}{4\pi \epsilon_o}q.\cfrac{Q}{r}$

$|\Delta P|=\cfrac{1}{4\pi \epsilon_o}q.\cfrac{Q}{xdx}-\cfrac{1}{4\pi \epsilon_o}\cfrac{qQ}{x+dx}$

$lim_dx \rightarrow 0=\cfrac{1}{4\pi \epsilon_o}.\cfrac{Q.2x}{x^2-(dx)^2}$

Where $dx=0$

so the answer is $x^2$

A rectangular surface of sides 10 cm and 15 cm is placed inside a uniform electric field of 25 V/m, such that the surface makes an angle

$30^{\circ}$ with the direction of electric field. Find the flux of the electric field through the rectangular surface

0%
$0.1675\ \ N m^2/C$
0%
$0.1875\ \ N m^2/C$
0%
$Zero$
0%
$0.1075\ \ N m^2/C$

Explanation

ATQ

$\vec { E } =25V/m$

$\vec { A } =10\times { 10 }^{ -2 }\times 15\times { 10 }^{ -2 }=1.5\times { 10 }^{ -2 }{ m }^{ 2 }$

Now, we know that

$\int { \phi } =\int { Eda }$

$=E.A$

$=Eacos\theta$

As the surface make

${ 30 }^{ 0 }$ but we have to calculate according to area vector which is perpendicular to the surface,

$\therefore \quad \rightarrow \theta ={ 90 }^{ 0 }-{ 30 }^{ 0 }={ 60 }^{ 0 }$

$\therefore \quad \phi =Eacos\theta =25\times 1.5\times { 10 }^{ -2 }\times cos{ 60 }^{ 0 }=0.1875\quad { Nm }^{ 2 }/C$

$\therefore$ Option (B) is the correct option.

Consider an electric field

$\bar { E } =E_{ 0 }\hat { x }$ where

$E_{0}$ is a constant. The flux through the shaded area (as shown in the figure) due to this field is

0%
$2E_{0}a^{2}$
0%
$\sqrt{2}E_{0}a^{2}$
0%
$E_{0}\ a^{2}$
0%
$\dfrac{E_{0}a^{2}}{\sqrt{2}}$

Explanation

on locating the plate is tilled lly

$45^o$

$\phi=E.S$

$=E_0a^2\cos 45$

$=\dfrac{E_0a^2}{\sqrt 2}$

1438001_1010537_ans_0385ed3167ae4d9187f89630f6a147fd.png

the point

$P$ , the flux of the electric field through the closed surface:

0%
will remain zero
0%
will becomes positive
0%
will become negative
0%
will become undefined

Explanation

We know that the flux of the electric field through the

$Q = EA\cos \theta$ Here, E is the electric field at point at which electric flux calculate the electric flux at P is undefined because the position of P w.r.t to closed surface is not defined

If the intensity of electric field at a distance

$x$ from the centre in axial position of small electric dipole is equal to the intensity at a distance

$y$ in equatorial position, then

0%
$x = y$
0%
$x = y/2$
0%
$y = x/2^{2/3}$
0%
$y = x/2^{1/3}$

Three charge +4q, Q and q are placed in a straight line of length $l$ at points distance 0, $\dfrac{l}{2}$ and $l$ respectively. What should be the value of Q in order to make the net force on q to be zero?

0%
$-q$
0%
$-2q$
0%
$-q/2$
0%
$4q$

Explanation

Two force are acting on

$q$ one due to

$4q$ and second due to

$Q$

$F_1=\cfrac{1}{4\pi\epsilon_0}\cfrac{4q^2}{12}\\F_2=\cfrac{1}{4\pi\epsilon_0}\cfrac{Qq}{(1/2)^2}\\ \quad=\cfrac{1}{4\pi\epsilon_0}\cfrac{(4Q)q}{12}\\F_1+F_2=0\\ \cfrac{1}{4\pi\epsilon_0}\cfrac{(4q)q}{12}+\cfrac{1}{4\pi\epsilon_0}\cfrac{(4Q)q}{12}=0\\Q=-q$

A solid insulating sphere of radius a carries a net positive charge 3Q, uniformly Distributed throughout its volume. Concentric with this sphere is a conducting Spherical shell with inner radius b and outer radius c and having a net charge Q, As shown in the figure.Electric field varies with distance r from the centre as $\left( {K = {1 \over {4\pi {\varepsilon _o}}}} \right)$

Explanation

$E = \dfrac{Kq_\epsilon}{r^2}$

when r < a,

$q_\epsilon \propto a, \\ E \propto a$

when r = a ,

$q_\epsilon = 3Q \\ E = \dfrac{K3Q}{a^2}$

when a < r < b ,

$q_\epsilon = constant, \, \, E\propto \dfrac{1}{r^2}$

when b < r < c ,

$E = 0$

when r > c ,

$q_\epsilon = 3 - 1 = 2Q , \, \, E = +ve$

Hence option 'C' is correct.

A charge situated at a certain from an electric dipole (small), in the end on position, experience a force

$F$ . Find the force acting on the same charge due to same dipole if its distance from the centre of dipole is doubled.

0%
$F/8$
0%
$F/4$
0%
$3F/8$
0%
$F/2$

Explanation

Force acting due to dipole at r distance

$=F\propto \dfrac 1{r^3}$

Centre getting double distance.

hence

$F'=\dfrac { F }{ 8 }$ .

A point charge Q is placed at origin O. Let $\overrightarrow {{E_A}}$ , $\overrightarrow {{E_B}}$ and $\overrightarrow {{E_C}}$ represent electric fields at A, B and C respectively. If coordination of A,B and C are respectively (1,2,3) m,(1,1,-1) m and (2,2,2) m then

0%
$\overrightarrow {{E_A}} \bot \overrightarrow {{E_B}}$
0%
$\overrightarrow {{E_A}} \parallel \overrightarrow {{E_B}}$
0%
$\left| {\overrightarrow {{E_B}} } \right|\parallel 4\left| {\overrightarrow {{E_C}} } \right|$
0%
$\left| {\overrightarrow {{E_B}} } \right|\parallel 8\left| {\overrightarrow {{E_C}} } \right|$

Explanation

$\overrightarrow{E_a}=\cfrac{h_E}{ra^3}=\cfrac{h_E}{1^2+2^2+3^2}(1\hat{i}+2\hat{j}+3\hat{k})\\ \overrightarrow{E_a}=\cfrac{h_E}{14^{3/2}}(1\hat{i}+2\hat{j}+3\hat{k})\\ \overrightarrow{E_b}=\cfrac{ha}{r_b^2}\overrightarrow{r_b}=\cfrac{h_E}{(1^2+1^2+1^2)}^{3/2}(1\hat{i}-1\hat{j}+1\hat{k})\\=\cfrac{h_E}{3^{3/2}}(1\hat{i}-1\hat{j}+1\hat{k})\\ \overrightarrow{E_c}=\cfrac{h_E}{r_c^2}\overrightarrow{r_c}=\cfrac{h_E}{(2^2+2^2+2^2)^{3/2}}(2\hat{i}+2\hat{j}+2\hat{k})$

$\quad=\cfrac{h_E}{12^{3/2}}(2\hat{i}+2\hat{j}+2\hat{k})$

Now

$\overrightarrow{E_a}.\overrightarrow{E_b}=\cfrac{h_E}{14^{3/2}}(1\hat{i}+2\hat{j}+3\hat{k})=\cfrac{h_E}{3^{3/2}}(1\hat{i}-1\hat{j}+1\hat{k})\\ \Rightarrow \overrightarrow{E_a}.\overrightarrow{E_b}=(\cfrac{h_E}{14^{3/2}})(\cfrac{h_E}{3^{3/2}})(1-2+3)\neq0$

Thus

$\overrightarrow{E_a}$ and

$\overrightarrow{E_b}$ are perpendicular to each other

$|E_c|=\cfrac{h_E}{12^{3/2}}(2^2+2^2+2^2)^{1/2}=\cfrac{h_E}{12^{3/2}}(12)^{1/2}\\ \Rightarrow |\overrightarrow{E_c}|=\cfrac{h_E}{12}\\ \overrightarrow|E_b|=\cfrac{ha}{3^{3/2}}(1^2+1^2+1^2)^{1/2}=\cfrac{_E}{3}=4\times\cfrac{h_E}{12}=4|\overrightarrow{E_c}|$

So,

$|\overrightarrow{E_b}|=4|E_c|$

A point charge is placed at the corner of a cube. The electric flux through the shaded surface is

0%
${q \over {8{\varepsilon _0}}}$
0%
${q \over {{\varepsilon _0}}}$
0%
${q \over {24{\varepsilon _0}}}$
0%
${q \over {12{\varepsilon _0}}}$

Explanation

$\textbf{Step 1: Choosing gaussian surface}$

To calculate the electric flux through the shaded area, we have to make a Gaussian surface such that a given charge appears at the center of the bigger cube resulting from the combination of

$8$ small cubes (as shown in figure)

$\textbf{Step 2: Flux through one cube}$

If a charge is at the corner then it is being shared by

$8$ cubes.

Flux passing through the given cube will be

$= \dfrac{q}{8\epsilon_0}$

$\textbf{Step 3: Electric flux through shaded area}$

Now for any cube, it is touching

$3$ of its faces. The area vector of that side and electric field vector will be perpendicular so flux through

$3$ sides will be

$0$ .

An equal amount of flux will flow from other

$3$ sides.

Flux through shaded area

$= \dfrac{q}{8 \times 3 \in_0}$

$= \dfrac{q}{24 \in_0}$

Hence, Option(C) is correct.

2111158_1019651_ans_97640c43dcee44a4b179db3f327cded9.png

A charge

$Q$ is divided into two parts of

$q$ and

$Q-q$ . If the Coulomb repulsion between them when they are separated, is to be maximum, the ratio of

$\dfrac{Q}{q}$ should be:

0%
$2$
0%
$1/2$
0%
$4$
0%
$1/4$

Explanation

Let parts are separated at a distance of 'd'

So repulsion force,

$F=\dfrac{kq(Q-q)}{d^2}$

here, Q and d are constant so for maximum force differentiate the force w.r.t q,

$\dfrac{dF}{dq}=\dfrac{k}{d^2}{(Q-2q)}$

For maximum force,

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

$\dfrac{k}{d^2}(Q-2q)=0$

$Q-2q=0$

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

Charge Q is uniformly distributed over a ring of radius r .Electric field at a point on the axis is maximum at a distance x from the centre. x is equal to

0%
$\dfrac{r}{\sqrt 2}$
0%
$\dfrac{r} {2}$
0%
$\sqrt 2 r$
0%
$r$

Explanation

The electric field due to the ring on its axis at a distance x is given by:-

$E = \dfrac{kqx}{(x^2 + R^2)^{3/2}}$

To find maximum electric field, we will use the concept of maximum and minimum :-

$\dfrac{dE}{dx} = kq \dfrac{(x^2 + R^2)^{3/2}-3/2 (x^2 + R^2 )^{1/2}.2x^2}{(x^2 + R^2)}$

Now,

$\dfrac{dE}{dx} = 0 \Rightarrow (x^2 + R^2)^{3/2} = \dfrac{3}{2} \times 2x^2 (x^2 + R^2)^{1/2}$

$\Rightarrow x^2 + R^2 = 3x^2$

$\Rightarrow 2x^2 = R^2$

$\Rightarrow x^2 = \dfrac{R^2}{2}$

$\Rightarrow x = \pm \dfrac{R}{\sqrt{2}}$

The total electric flux through a cube when a charge

$8q$ is placed at one corner of the cube is

0%
${ \varepsilon }_{ 0 }q$
0%
$\dfrac { { \varepsilon }_{ 0 } }{ q }$
0%
$4\pi { \varepsilon }_{ 0 }q$
0%
$\dfrac { q }{ { \varepsilon }_{ 0 } }$

A square of side 'a' is lying in xy plane that two of its sides are lying on the axis. If

$\vec{E} = E_0 x \hat{k}$ is applied on the square. The flux passing through the square is

0%
$E_0 a^2$
0%
$\dfrac{E_0 a^3}{2}$
0%
$\dfrac{E_0 a^3}{3}$
0%
$\dfrac{E_0 a^2}{2}$

Explanation

Square side

$=a$

$\overline { E } ={ E }_{ 0 }x\hat { k }$

$\phi =\overline { E } .\overline { A }$

$=\left( { E }_{ 0 }x\hat { k } \right) .\left( { a }^{ 2 } \right)$

$={ E }_{ 0 }{ a }^{ 2 }$

The ratio of the electric force between two proton to that between two electrons under similar conditions is the order of:

0%
$10^{42}$
0%
$10^{39}$
0%
$10^{36}$
0%
$1$

Explanation

${F_{electric}} = \dfrac{{k{q_1}{q_2}}}{{{r^2}}}$

$\alpha \,\,\,$

$\,{q_1}{q_2}$

$\dfrac{{{F_{electric}}}}{{{F_{proton}}}} = 1$

Hence,

option

$(D)$ is correct answer.

A point charge

$+q$ is placed at the centre of a cube of side

$L$ . The electric flux emerging from the cube is

0%
$\cfrac{q}{{ \varepsilon }_{ 0 }}$
0%
zero
0%
$\cfrac{6q{L}^{2}}{{ \varepsilon }_{ 0 }}$
0%
$\cfrac{q}{6{L}^{2}{ \varepsilon }_{ 0 }}$

Explanation

Hint: Electric Flux linked with any surface is the count of Electric Field lines passing through that surface.

Correct Option: A.

Explanation for the correct answer:

$\bullet$ It is given that a charge

$+q$ is placed at the center of the cube of the side

$L$ . Thus, we can say that a charge is placed inside a closed surface, that is the cube.

$\bullet$ For closed surface, Gauss Law of Electrostatics states that Electric Flux linked with any closed surface is always

$\dfrac{1}{\epsilon_0}$ times the total charge enclosed by that surface.

$\bullet$ In our question, the cube is a closed surface, and it encloses the charge

$+q$ . Thus, Electric Flux linked with the cube can be given as

$\phi = \dfrac{1}{\epsilon_0} \times (+q)$

Or,

$\phi = \dfrac{+q}{\epsilon_0}$

Thus, Electric Flux emerging from cube is given as, $\phi = \dfrac{+q}{\epsilon_0}$ .

The chasrge density of an insulating infinity surface is $\left( {e/\pi } \right)\;c{m^2}$ then the field intensity at a nearby point in volt/meter will be -

0%
$2.88 \times {10^{ - 12}}$
0%
$2.88 \times {10^{ - 10}}$
0%
$2.88 \times {10^{ - 9}}$
0%
$2.88 \times {10^{ - 18}}$

Which of the following aren't the properties of lines of force of an electric field?

Lines of force always intersect.
Lines of force start from negative charge and terminate at te positive charge.
The tangent to line of force at a point gives the direction of the electric field.
Each unit positive charge gives rise to $1 / 4 \pi \epsilon_0$ lines of force in free space.

0%
1 and 2
0%
2 and 4
0%
1 and 3
0%
2 and 3

Explanation

1.The electric lines of force are imaginary lines the direction of filed lines for the positive charge is away from the centre and for the negative charge the field lines move towards the centre.

2. The field lines can never intersect each other because at the intersection point there can be two direction of field which is not possible.

3. The field originated from positive charge and terminate at the negative charge.

4. The lines of the force in free space is given by $E = \frac{q}{{4\pi {\varepsilon _0}{r^2}}}$

A charged particle

$q$ is projected in an electric field produced by a fixed point charge

$Q$ as shown in the figure. Select the correct statements:

0%
The path taken by $q$ is a straight line
0%
The path taken by $q$ is not a straight line
0%
The minimum distance between the two particle is $\dfrac{\dfrac{qQ}{2\pi\varepsilon_{0}}+\sqrt{\left(\dfrac{qQ}{2\pi\varepsilon_{0}}\right)^{2}+4m^{2}u^{2}a^{2}}}{2mu^{2}}$
0%
Velocity of the particle $q$ is changing in both magnitude and direction.

The work done by an external force in rotating dipole of dipole moment

$\vec{P}$ slowly in uniform electric field of intensity

$\vec{E}$ from

$0^0$ to

$180^0$

0%
$\text{2 PE}$
0%
$\text{8 PE}$
0%
$\text{9 PE}$
0%
$\text{5 PE}$

Explanation

Work done by an external force in rotating dipole of dipole moment

$=\overline { P }$

Uniform electric field of intensity

$\overline { E }$ from

${ 0 }^{ 0 }$ to

${ 180 }^{ 0 }$

The work done is given by

$dw=T\times d\left( \theta \right)$

Torque at any particular angle

$T.E\times q\times d\sin\theta$

total work done

$=\left[ INT \right] \left( dw \right)$

$=\left[ INT \right] \left( E\times d\times q\sin\theta \right) d\left( \theta \right)$

$=Eqd\left( -1-1 \right)$

$=2Eqd$ will be the required

$=2PE$ (

$\because$

$qd=p$ )

A charged oil drop od mass

$2.5\times { 10 }^{ -7 } kg$ is in space between the two plates, each of area

$2\times { 10 }^{ -2 } { m }^{ 2 }$ of a parallel plate capacitor. When the upper plate has a charge of

$5\times { 10 }^{ -7 } C$ and the lower plate has an equal negative charge then the oil remains stationery. The charge of the oil drop is (take,

$g=10 { m/s }^{ 2 }$ )

0%
$9\times { 10 }^{ -1 } C$
0%
$9\times { 10 }^{ -6 } C$
0%
$8.85\times { 10 }^{ -13 } C$
0%
$1.8\times { 10 }^{ -14 } C$

Explanation

Mass

$=2.5\times { 10 }^{ -7 }kg$

$qE=mg$

or,

$q\times \dfrac { Q }{ A{ \epsilon }_{ 0 } } =mg$

$q=\dfrac { mgA{ \epsilon }_{ 0 } }{ Q }$

$=\dfrac { 2.5\times { 10 }^{ -2 }\times 2\times { 10 }^{ -2 }\times 8.854\times { 10 }^{ -12 }\times 10 }{ 5\times C7\times 1 } =9\times { 10 }^{ -14 }C$

1236223_1029533_ans_ea9ba893d80b4f51bd543a539c758ea9.jpg

What is the angle between the electric dipole moment and the electric field strength due to it on the equatorial line?

0%
$0^o$
0%
$90^o$
0%
$180^o$
0%
none of these

Explanation

$\textbf{Step 1: Angle between Electric field and Dipole}$

$\textbf{[Refer Fig.]}$

At a point

$P$ on equatorial line

$\vec{E}_{R}$ is net electric field due to dipole at point P as shown in the figure, directed Leftwards.

The direction of dipole vector

$\vec P$ is from

$-q$ to

$+q$ charge, i.e. Rightwards.

Therefore, The angle between electric dipole moment and electric filed on the equitorial line is

$180^{\circ}$

Hence option

$(C)$ correct

2114281_1061648_ans_a2d117d2a2444ff6b0695c169b0e61e4.png

The line integral of an electric field along the circumference of a circle of radius r drawn with a point Q at the centre will be _____

0%
$\dfrac{1}{4 \pi^2 to } \dfrac{Q}{r}$
0%
$\dfrac{Q}{2 \pi \varepsilon_0 r}$
0%
Zero
0%
$2 \pi Q r$

Explanation

$\begin{array}{l} To\, calculate \\ \oint { \overrightarrow { E } .\overrightarrow { dl } } \\ \oint { \overrightarrow { E } .\overrightarrow { dl } } \, is\, zero\, everywhere\, along\, the\, { { int } }ergral\, will\, be\, zero. \\ \sin ce, \\ \overrightarrow { E } \, and\overrightarrow { dl } \, are\, perpendicular\, to\, each\, other\, \\ \overrightarrow { E } .\overrightarrow { dl } =0 \\ thus \\ \oint { \overrightarrow { E } .\overrightarrow { dl } } =0 \\ Hence,\, the\, correct\, option\, is\, C. \end{array}$
1226641_1048441_ans_377349add18d441888242657b10c265d.PNG

1226641_1048441_ans_377349add18d441888242657b10c265d.PNG

Shown below is a distribution of charges. The flux of electric field due to these charges through the surface S is

0%
${3q/\epsilon_o}$
0%
${2q/\epsilon_o}$
0%
${q/\epsilon_o}$
0%
zero

Explanation

Total net flux

$=0$
But flux will be present in the region of surface

$S$

1174500_1070813_ans_ca4f620ef7384fb0be061598f4cb75cc.png

At a certain distance from a point charge the electric field is

$500\ V/m$ and the potential is

$3000\ V$ . What is this distance:

0%
$36\ m$
0%
$12\ m$
0%
$6\ m$
0%
$144\ m$

Explanation

Given that,

Electric field $E=500\,V/m$

Potential $V=3000V$

We know that,

$E=\dfrac{V}{d}$

$d=\dfrac{V}{E}$

$d=\dfrac{3000}{500}$

$d=6\,m$

Hence, the distance is $6\ m$

In the figure shown here, A is a conducting sphere and B is a closed spherical surface. If a-q change is placed at C near A, then the electric flux through the closed surface is -

0%
zero
0%
positive
0%
negative
0%
none of the above can be predicted

Explanation

As the electric field inside the spherical conductor has to be zero the equal and opposite charges will be distributed on the conductor surface such that the induced electric field counters the electric field due to negative charge. For negative charges will be distributed on right and positive on left side

As the spherical region is on right side

$q$ inside

$< 0$ .

So, by Electric there

$=\dfrac {g\ inside}{E_0} < 0$

Option

$C$ is correct.

1418333_1083017_ans_229fbf6907ff4124b0beef0188ba4a07.png

An electron (of charge

$-e$ ) revolves around a long wire with uniform charge density

$\lambda$ in a circular path of radius

$r$ . Its kinetic energy is given by:

0%
$\dfrac { \lambda e }{ 8\pi { \epsilon }_{ 0 }r }$
0%
$\dfrac { \lambda e }{ 4\pi { \epsilon }_{ 0 }r }$
0%
$\dfrac { \lambda e }{ 2\pi { \epsilon }_{ 0 }r}$
0%
$\dfrac { \lambda e }{ 4\pi { \epsilon }_{ 0 } }$

Explanation

Electric field for infinitely long wire is given by $E=\dfrac{\lambda }{2\pi {{\varepsilon }_{0}}r}\,where\,\,\lambda =ch\arg e\,\,density$

Since the electron is revolving then the centripetal force is balanced by the electrostatic force

${{F}_{e}}={{F}_{c}}$

$eE=\dfrac{m{{v}^{2}}}{r}$

$\dfrac{e\lambda }{2\pi {{\varepsilon }_{0}}}=m{{v}^{2}}$

$\dfrac{e\lambda }{4\pi {{\varepsilon }_{0}}}=\dfrac{1}{2}m{{v}^{2}}$

$KE=\dfrac{e\lambda }{4\pi {{\varepsilon }_{0}}}$

The figure shows a family of equipotential surfaces and four paths along which an electron is made to move from one surface to another as shown in figure.
1) What is the direction of electric field?
2) Rank the paths according to the magnitude of the work done, greatest first

0%
$Rightward ; 4 > 3 > 2 > 1$
0%
$Leftward ; 1 > 2 > 3 > 4$
0%
$Rightward ; 3 = 4 > 2 = 1$
0%
$Leftward ; 1 > 2 > 3 = 4$

Explanation

Along the direction of electric field, electric potential decreases, hence it should be from left to right.

Change in potential of path 3 and 4 is greater hence

$3= 4 > 2 =1$

A plane surface of area

$10\ cm^{2}$ is placed in a uniform electric field of

$20\ N/C$ such that the angle between the surface and the electric field is

$30^{o}$ . The electric flux over the surface is:

0%
$0.0173\ Vm$
0%
$0.0236\ Vm$
0%
$0.01\ Vm$
0%
$0.0346\ Vm$

Explanation

Given,

Electric field, $E=20\,N{{C}^{-1}}$

Area, $A=10\times {{10}^{-4}}\,{{m}^{2}}$ A

Electric Flux $\phi =E.A\,\cos \theta =20\times 10\times {{10}^{-4}}\cos {{60}^{o}}=0.01\,Vm$

1023889_1082552_ans_f05b3b447f8e44c989b4bdf21a5cd781.jpg

A point charge of value

$10^{-7}\ C$ is situated at the centre of cube of

$1\ m$ side. The electric flux through its total surface area is:

0%
$113\times 10^{4}\ Nm^{2}/C$
0%
$11.3\times 10^{4}\ Nm^{2}/C$
0%
$1.13\times 10^{4}\ Nm^{2}/C$
0%
$none\ of\ these$

Explanation

Given that,

Charge $q={{10}^{-7}}\,C$

We know that,

The electric flux is

$\phi =\dfrac{q}{{{\varepsilon }_{0}}}$

$\phi =\dfrac{{{10}^{-7}}}{8.85\times {{10}^{-12}}}$

$\phi =0.1129\times {{10}^{5}}$

$\phi =1.13\times {{10}^{4}}\,N{{m}^{2}}/C$

Hence, the flux is $1.13\times {{10}^{4}}\,N{{m}^{2}}/C$

A wire of linear charge density

$\lambda$ passes through a cuboid of length

$\ell$ , breadth

$b$ and height

$h\ (\ell>b>h)$ in such a manner that flux through the cuboid is maximum. The position of the wire is now changed, so that the flux through the cuboid is minimum. The ratio of maximum flux to minimum flux will be :

0%
$\dfrac{\sqrt{l^{2}+b^{2}}}{h}$
0%
$\dfrac{\sqrt{l^{2}+b^{2}+h^{2}}}{h}$
0%
$\dfrac{h}{\sqrt{l^{2}+b^{2}}}$
0%
$\dfrac{l}{\sqrt{l^{2}+b^{2}+h^{2}}}$

Explanation

Formula,

$Q=\dfrac{q}{\epsilon _0}$

maximum flux occurs diagonally,

$Q_{max}=\dfrac{\sqrt{L^2+B^2+H^2}}{\epsilon _0}$

minimum flux occurs when height equals length,

$Q_{min}=\dfrac{H}{ \epsilon _0}$

the ratio of maximum by minimum flux is given by,

$\dfrac{Q_{max}}{Q_{min}}=\dfrac{\dfrac{\sqrt{L^2+B^2+H^2}}{\epsilon _0}}{\dfrac{H}{ \epsilon _0}}$

$\therefore \dfrac{Q_{max}}{Q_{min}}=\dfrac{\sqrt{L^2+B^2+H^2}}{H}$

A point charge of

$+6\mu C$ is placed at a distance 20 cm directly above the centre of a square of side 40 cm. The magnitude of the flux through the square is

0%
$\epsilon_0$
0%
$\dfrac{1}{\epsilon_0}$
0%
$\epsilon_0\times 10^{-6}$
0%
$\dfrac{1}{\epsilon_0} \times 10^{-6}$

Explanation

Given,

$Q_{en}=+6\mu C=+6\times 10^{-6}C$

From the Gauss's law, the magnetic flux through the cube is given by

$\phi=\dfrac{Q_{en}}{\varepsilon_0}$

The magnetic flux through the one face of the cube or square is,

$\phi_s=\dfrac{Q_{en}}{6 \varepsilon_0}$

$\phi_s=\dfrac{6\times 10^{-6}}{6\varepsilon_0}$

$\phi_s=\dfrac{1}{\varepsilon_0}\times 10^{-6}$

The correct option is D.

If the dipole is place in a non-uniform electric field an angle

$\theta$ , in addition tarque

0%
Experiences a force
0%
Experiences a repulsive force
0%
No any kind of force
0%
Neither attractive nor repulsive force

Explanation

When a electric dipole is held in a non-uniform electric field then electric dipole will experience a torque when dipole vector is not parallel to electric field direction. Then torque

$\tau =P\varepsilon \sin\theta.$ Where

$P$ and

$\varepsilon$ are parallel and angle

$\theta$ between two vector is zero hence there will not be any torque. So torque experience force.

Hence, the answer is experiences a force.

There are two charges

$+2\mu\ C\ and -3\mu\ C$ . The ratio of forces acting on them will be

0%
$2:3$
0%
$1:1$
0%
$3:2$
0%
$4:9$

A charge q is to be distributed on two conducting spheres. What should be the value of the charges on the spheres so that the repulsive force between them is maximum when they are placed at a fixed distance from each other in air?

0%
$\dfrac{q}{2}$ and $\dfrac{q}{2}$
0%
$\dfrac{q}{4}$ and $\dfrac{3q}{4}$
0%
$\dfrac{q}{3}$ and $\dfrac{2q}{3}$
0%
$\dfrac{q}{5}$ and $\dfrac{4q}{5}$

Explanation

Suppose Q on one sphere &

q - Q on other sphere

total charge is still

$a+q-Q=q$

placed at distance r (constant)

then force =

$\dfrac{1}{4\pi c_{10}}\dfrac{Q(q- Q)}{r^{2}}$

$F= K$

$Q(q-Q)$

$ik= \dfrac{1}{4\pi c_{10}r^{2}}$

$\dfrac{dF}{dQ}= \dfrac{kd}{dQ}\left \langle Qq-Q^{2} \right \rangle= k \left \langle q-2Q \right \rangle$

For maximum

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

$\Rightarrow q-2Q=0\Rightarrow Q= q/2$

$F_{max}= \dfrac{1}{4\pi c_{10}}\dfrac{\dfrac{q}{2}.\dfrac{q}{2}}{r^{2}}$

i.e. For charge values

$\dfrac{q}{2}$ &

$\dfrac{q}{2}$

CBSE Questions for Class 12 Medical Physics Electric Charges And Fields Quiz 8 - MCQExams.com

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Practice Class 12 Medical Physics Quiz Questions and Answers

CBSE Questions for Class 12 Medical Physics Electric Charges And Fields Quiz 8 - MCQExams.com

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Practice Class 12 Medical Physics Quiz Questions and Answers

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