MCQ Questions for Class 12 Medical Physics Electric Charges And Fields Quiz 5

Identify the correct statement about the charges

$q_1$ and

$q_2$ :

0%
$q_1$ and $q_2$ , both are positive
0%
$q_1$ and $q_2$ , both are negative
0%
$q_1$ is positive and $q_2$ is negative
0%
$q_2$ is positive and $q_1$ is negative

Explanation

The filed of lines is outward from the positive charge and inward to the negative charge.
Here the field of lines is towards the charges

$q_1$ and

$q_2$ . Thus, both charges are negative.

Units of electric flux are :

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$\displaystyle \dfrac{N-m^2}{C^2}$
0%
$\displaystyle \dfrac{N}{C^2 - m^2}$
0%
$volt - m$
0%
$volt - m^3$

Explanation

Electric flux

$=$ electric field

$\times$ area

$=|dV/dx| \times m^2=V/m \times m^2=V.m$
or Electric flux

$=$ electric field

$\times$ area

$=F/q \times A=N/C \times m^2=\dfrac{N.m^2}{C}$

The electric field in a region of space is given by

$\vec{E} = 5\widehat{i} + 2\widehat{j} \ N/C$ . The flux of

$\vec{E}$ due to this field through an area

$1m^2$ lying in the y-z plane, in SI units, is :

0%
$5$
0%
$10$
0%
$2$
0%
$5 \sqrt{29}$

Explanation

The flux ,

$\phi=\vec{E}.\hat{n}da=(5\hat{i}+2\hat{j}).\hat{i}(1)=5$ as area y-z plane so normal vector is along x direction.

Four dipoles each one having magnitudes of charges

$\pm$ e are placed inside a sphere. The total flux of

$\vec{E}$ coming out of the sphere is :

0%
Zero
0%
$\displaystyle \dfrac{4 e}{\varepsilon_0}$
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$\displaystyle \dfrac{8 e}{\varepsilon_0}$
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None of these

Explanation

Each dipole has equal and opposite charge. So net charge of dipole is zero.

Thus the total charge inside the sphere is zero. According to Gauss's law, the flux, $\phi=\dfrac{q_{in}}{\epsilon_0}=0$ as charge inside sphere is zero.

$5$ charges each of magnitude

$10^{-5}$

$C$ and mass

$1 kg$ are placed (fixed) symmetrically about a movable central charges of magnitude

$5 \times 10^{-5}C$ and mass

$0.5 kg$ as shown. The charge at

$P_{1}$ , is removed. The acceleration of the central charge is

$[$ Given

$OP_1 = OP_2 = OP_3 = OP_4 = OP_5 = 1m;$

$\displaystyle \frac{1}{4\pi \varepsilon _{0}}=9\times 10^{9}$ in

$SI$ units

$]$

0%
$9 m s^{-2}$ upwards
0%
$9 m s^{-2}$ downwards
0%
$4.5 m s^{-2}$ upwards
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$4.5 m s^{-2}$ downwards

Explanation

Now Force on Q = Force in figure (1) + Force in fugure (2)
F =

$0\quad +\quad \frac { KqQ }{ { R }^{ 2 } } \quad \uparrow upward\\$
F =

$0\quad +\quad \frac { \left( 9\times { 10 }^{ 9 } \right) \times { 10 }^{ -5 }\times \left( 5\times { 10 }^{ -5 } \right) }{ 1 } \\$

$ma\quad =\quad 9\times { 10 }^{ -1 }\times 5\\ 0.5a\quad =\quad 9\times { 10 }^{ -1 }\times 5\\ a\quad =\quad 9\frac { m }{ { sec }^{ 2 } } \quad \uparrow upward$

Two small objects each with a net charge of +Q exert a force of magnitude F on each other. We replace one of the objects with another whose net charge is + 4Q. We move the +Q and +4Q charges to be 3 times as far apart as they were. What is the magnitude of the force on the +4Q charge ?

0%
F
0%
4F
0%
4F/3
0%
4F/9
0%
F/3

Explanation

Initially

$q_1=q_2=Q$ and

$r=R$

Force was

$F=\dfrac{q_1q_2}{4\pi\epsilon_o r^2}=\dfrac{Q^2}{4\pi\epsilon_o R^2}$

Thenafter:-

$q_1=Q, q_2=4Q$ and

$r=3R$

Force on charge 4Q,

$F'=\dfrac{Q.4Q}{4\pi\epsilon_o (3R)^2}=\dfrac{4Q^2}{4\pi\epsilon_o (9R^2)}$

$\implies F'=\dfrac{4}{9}F$

Answer-(D)

Two charges of

$+1$

$\mu C$

$\&+5$

$\mu C$ are placed

$4 cm$ apart, the ratio of the force exerted by both charges on each other will be -

0%
$1:1$
0%
$1:5$
0%
$5:1$
0%
$25:1$

Explanation

From Newton's Third law of motion , force exerted by charge

$1\mu C$ on

charge

$5\mu C$ is equal to the force by charge

$5\mu C$ on charge

$1\mu c$ in magnitude.

Hence ratio of force is

$1:1$

Answer-(A)

In winter season, a mild spark is often seen when a man touches somebody else's skin. Why?

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Due to lack of humidity and rubbing with clothes, charge accumulate on human body which is discharged via sparking
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Due to cold, electrostatic charge on body finds a lower resistance path to the skin of other's body
0%
The static charge on sweaters worn by the two persons is different, hence discharge through sparking occurs
0%
Similar to the lightning, extremely high potential exists on both the bodies and hence they discharge through sparking

The electric potential due to an infinite sheet of positive charge density

$\sigma$ at a point located at a perpendicular distance

$Z$ from the sheet is (Assume

${V}_{0}$ to be the potential at the surface of sheet) :

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${V}_{0}$
0%
${V}_{0}-\cfrac{\sigma Z}{{\epsilon}_{0}}$
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${V}_{0}+\cfrac{\sigma Z}{2{\epsilon}_{0}}$
0%
${V}_{0}-\cfrac{\sigma Z}{2{\epsilon}_{0}}$

Explanation

$\int { dV } =-\int { E.dr } \Rightarrow \quad \int _{ { V }_{ 0 } }^{ V }{ dV } =-\int _{ 0 }^{ Z }{ \dfrac { \sigma }{ 2{ \in }_{ 0 } } dZ }$

$\Rightarrow V-{ V }_{ 0 }=-\dfrac { \sigma Z }{ 2{ \in }_{ 0 } } \Rightarrow V={ V }_{ 0 }-\dfrac { \sigma Z }{ 2{ \in }_{ 0 } }$

Two identical charges of magnitude +Q are fixed as shown . A third charge -Q is placed mid way between them at point P. Then small displacements of -Q are made in the directions indicated by arrows. The -Q is stable with respect to displacement .

0%
I and III
0%
I and V
0%
II and $IV$
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III and V
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stable for any small displacement.

The figure given below shows a positively charged electroscope. In fig.(a), the deflection of leaves is shown. Now, you touch the metal sphere of the electroscope (see fig.b). What will be your observation?

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Leaves will achieve the vertical position (uncharged position)
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No change will take place in their deflection
0%
Deflection will become more
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Deflection may be more or less depending on the amount of charges present on the leaves

Two point charges

$q_1$ and

$q_2$ are placed at a distance of 50 m from each other in air, and interact with a certain force. The same charges are now put in oil whose relative permittivity isIf the interacting force between them is still the same, their separation now is :

0%
16.6 m
0%
22.3 m
0%
28.0 m
0%
25.0 cm

Explanation

Here, $F_{air}=\dfrac{q_1q_2}{4\pi\epsilon_0 d^2_{air}}$ and

$F_{oil}=\dfrac{q_1q_2}{4\pi\epsilon_r\epsilon_0 d^2_{oil}}$

According to question, $F_{air}=F_{oil}$

or $\dfrac{q_1q_2}{4\pi\epsilon_0 d^2_{air}}=\dfrac{q_1q_2}{4\pi\epsilon_r\epsilon_0 d^2_{oil}}$

or $\epsilon_r d^2_{oil}=d^2_{air}$

or $d^2_{oil}=\dfrac{(50)^2}{5}=500$

or $d_{oil}=22.3 m$

An electric dipole is placed perpendicular to an infinite line of charge at some distance as shown in figure. Identify the correct statement.

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The dipole is attracted towards the line charge
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The dipole is repelled away from the line charge
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The dipole does not experience a force
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The dipole experiences a force as well as a torque

In the given figure calculate the force on unit positive charge placed at

$A$ . The length of each side on the square is

$4 cm$ .

0%
$\displaystyle 9\sqrt{3} \times 10^{4} dyn$
0%
$\displaystyle 9\sqrt{3} \times 10^{-4} dyn$
0%
$\displaystyle 9\sqrt{3} \times 10^{4} N$
0%
$\displaystyle 9\sqrt{3} \times 10^{-4} N$

When electrons are added to an uncharged body, then the body

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gets negatively charged
0%
gets positively charged
0%
remains uncharged
0%
gets negatively or positively charged depending upon its size

In Coulomb's law, the constant of proportionality K has the units

$Nm^2/C^2$ then the magnitude of K in air is :

0%
$9 \times 10^5$
0%
$9 \times 10^{9}$
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$9 \times 10^3$
0%
none of these

Explanation

In Coulomb's law, the value of constant proportionality is

$K=\dfrac{1}{4\pi\epsilon_0}=\dfrac{1}{4\times 3.14\times (8.85\times 10^{-12})}=8.99\times 10^{9} \sim 9\times 10^9 Nm^2/C^2$

In Coulomb's law, the constant of proportionality K has the units :

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$N$
0%
$N^2$
0%
$NC^2/m^2$
0%
$Nm^2/C^2$

Explanation

By Coulomb's law, the force between the charges is

$F=\dfrac{Kq_1q_2}{r^2}$ where

$r$ is the distance between them.

So,

$K=\dfrac{Fr^2}{q_1q_2}=\dfrac{Nm^2}{C^2}$

(Unit of force

$F$ is

$N$ , unit of distance is

$m$ and unit of charge is

$C$ )

Two identical metallic spheres are given charges

$+q$ and

$-q$ respectively now :

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Both sphere have equal masses
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The positively charged sphere has a mass smaller than the negatively charged sphere
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The change in their masses depends on the magnitude of $q$
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Both (A) & (C)

$12$ positive charges of magnitude

$q$ are placed on a circle of radius

$R$ in a manner that they are equally spaced.

$A$ charge

$+Q$ is placed at the centre. If one of the charges

$q$ is removed, then the force on

$Q$ is :

0%
Zero
0%
$\displaystyle \frac {qQ}{4\pi \varepsilon _{0}R^{2}}$ away from the position of the removed charge.
0%
$\displaystyle \frac {11qQ}{4\pi \varepsilon _{0}R^{2}}$ away from the position of the removed charge.
0%
$\displaystyle \frac {qQ}{4\pi \varepsilon _{0}R^{2}}$ towards the position of the removed charge.

Explanation

Now force on charge Q = force in figure (i) + force in figure (ii)
F =

$0\quad +\quad \frac { KqQ }{ { R }^{ 2 } }$
F =

$\frac { KqQ }{ { R }^{ 2 } }$ towards the position of that removed charge.

Two identical spheres having charges

$8 \mu C$ and

$-4\mu C$ are kept at a certain distance apart. Now they are brought into contact, after that again they are kept at the same distance. Compare the forces in the two cases

0%
$4 : 1$
0%
$12 : 1$
0%
$8 : 1$
0%
$16 : 1$

Explanation

If two identical sphere having charge

$8\: \mu C$ and

$-4\: \mu C$ are kept a certain distance apart, then force between them is

$F$ .

Let the distance between charged sphere is

$d$ .

then,

$F=\dfrac{k\times 8\times (-4)\times 10^{-12}}{d^{2}}$ [attractive]

$d^{2}=\dfrac{32k\times 10^{12}}{F}$ ....(i)

Now, when both the identical sphere touch, then charge on each sphere is average of initial charge on sphere.

i.e,

$Q=\dfrac{(8-4)\:\mu C}{2}= 2\; \mu C$ on each sphere.

Now, Force between them is:

$F'=\dfrac{k\times (2\times 10^{-6})^{2}}{d^{2}}$

From equation (i), we get

$F'=\dfrac{k\times 4\times 10^{-12}\times F}{32k\times 10^{-12}}$

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

so, the ratio is

$8:1$

Opening up of strips of an electroscope is an indication of:

0%
Proximity of a conductor.
0%
Proximity of a charged body.
0%
Proximity of an uncharged body.
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Earth connection.

The figure shows two situations in which a Gaussian cube is placed in an electric field. The arrows and values indicate the directions and magnitudes (in

$\displaystyle N-{ { m }^{ 2 } }/{ C }$ ) of the electric fields.What is the net charge (in the two situations) inside the cube?

0%
(1) negative, (2) positive
0%
(1) negative, (2) zero
0%
(1) positive , (2) positive
0%
(l) negative, (2) negative

Explanation

Given the 2 situation in which a Gaussian cube sits in the electric field. the magnitudes & direction of field line are given in the figure we have to find net charge in both situation.

Hear we use Gauss's law,

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

so,

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

q=charge enclosed within the Gaussian surface.

${ \varepsilon }_{ 0 }=$ permeability in air.

$\phi =flux\quad$ through a surface.

$\phi =$ electric field

$\times$ area of Gaussian surface.

since area of Gaussian surface is same in both the case;so the net charge depend on the net flux through surface.

flux in case(1) : lines coming out -lines entering.

$=6+7-2-15-7-8$

$=-ve$

so; charge is (1) situation is negative.

in case (2), net flux

$=9+3-2-6-7-5$

$=-ve$

so charge in

$-ve$ in case(2)

so; the answer is both negative charge.

Two point charges +q and -q are held fixed at (-d, 0) and (d, 0) respectively of a x - y coordinate system. Then :

0%
The electric field E at all points on the axis has the same direction
0%
Work has to be done in bringing a test charge from $\displaystyle \infty$ to the orgin
0%
Electric field at all points on y-axis is along x-axis
0%
The dipole moment is 2qd along the x-axis

Explanation

If we take a point M on the X-axis as shown in the figure, then the net electric field is in X-direction.

∴ Option (a) is incorrect.

If we take a point N on Y-axis, we find net electric field along +X direction. The same will be true for any point on X-axis.

So, (c) is a correct option.

The work is calculated as:

$W_{\infty to 0} = q(V_{\infty} - V_{0}) = q(0-0) = 0$

(b) is incorrect.

The direction of the dipole moment is from-:-ve to +ve and it is not along the X-axis. Therefore (d) is incorrect.

When a comb is rubbed with hair, it attracts paper bits. Choose the right explanation:

0%
Bits of paper gets attracted due to gravitational force
0%
Due to electromagnetic effect, bits of paper are attracted
0%
Comb gets charged by friction and attracts bits of paper
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None of these

Two point charges

$+3\mu C$ and

$+8\mu C$ repel each other with a force of

$40 N$ . If a charge of

$-5\mu C$ is added to each of them, the force between them will become:

0%
$-10 N$
0%
$10 N$
0%
$20 N$
0%
$-20 N$

Explanation

Redistribution of charges takes place.
Charge

${ q }_{ 1 }=3\mu C$
Charge

${ q }_{ 2 }=8\mu C$
When third charge

${ q }_{ 3 }=-5\mu C$ is added to each, then new charges on

${ q }_{ 1 }$ and

${ q }_{ 2 }$ will be

${ q }_{ 1 }^{ \prime }=3-5=-2\mu C$
and,

${ q }_{ 2 }^{ \prime }=8-5=3\mu C$
Now,
In first case,

$40=\dfrac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \cdot \dfrac { 3\times 8 }{ { r }^{ 2 } }$
In second case,

$F=\dfrac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \times \dfrac { \left( -2\times 3 \right) }{ { r }^{ 2 } }$

$\therefore \dfrac { F }{ 40 } =\dfrac { -2\times 3 }{ 3\times 8 }$
or

$F=-10N$

A dipole of dipole moment 'p' is placed in non-uniform electric field along x-axis. Electric field is increasing at the rate of

$1\ V m^{-1}$ . The force on dipole is:

0%
$0$
0%
$2p$
0%
$p/2$
0%
$p$

Explanation

Force is given as the negative gradient of potential energy.

$F=\dfrac { -dU }{ dx }$

Potential energy for an electric dipole is

$U=-\vec { p } .\vec { E }$

So, we get the force as,

$F=p\dfrac { dE }{ dx } =p$
As

$\dfrac { dE }{ dx } =1$

Which of the following represents Coulomb's constant?

0%
$k\,=\,\frac{1}{4\times \pi \times e^0}$
0%
$k\,=\,1$ + ( $4\times \pi \times e^0$ )
0%
$k\,=\,1$ - ( $4\times \pi \times e^0$ )
0%
$k\,=\,1$ * ( $4\times \pi \times e^0$ )

Explanation

We know that Coulomb's force,

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

The Coulomb constant will be

$k=\frac{1}{4\pi\epsilon_0}$

What is known as electrostatic repulsion?

0%
When two positively or two negatively charged particles come closer to each other then they repel from each other
0%
When two charged particles come closer to each other then they repel from each other
0%
When positively and negatively charged particles come closer to each other then they repel from each other
0%
None

What does coulomb's law of attraction and repulsion states?

0%
F is directly proportional to $q_1q_2$ and inversely proportional to $(r_{12})^2$
0%
F is directly proportional to $q_1q_2$ and inversely proportional to $(r_{12})$
0%
F is indirectly proportional to $q_1q_2$ and directly proportional to $(r_{12})^2$
0%
None

Explanation

The Coulomb law is the force between two charges

$q_1$ and

$q_2$ when the separation between them is

$r_{12}$ .

Here ,

$F=\dfrac{kq_1q_2}{r_{12}^2}$

Thus, F is proportional to

$q_1q_2$ and invewrsely proportional to

$r_{12}^2$ .

The line AA' is a charged infinite conducting plane which is perpendicular to the plane of the paper. The plane has a surface density of charge

$\sigma$ and B is ball of mass m with a like charge of magnitude q. B is connected by string from a point on the line AA' The tangent of angle (

$\theta$ ) formed between the line AA' and the string is:

0%
$\displaystyle \frac{q\sigma}{2\varepsilon_0mg}$
0%
$\displaystyle \frac{q\sigma}{2\pi \varepsilon_0mg}$
0%
$\displaystyle \frac{q\sigma}{6\pi \varepsilon_0mg}$
0%
$\displaystyle \frac{q\sigma}{\varepsilon_0mg}$

Explanation

The electrostatic force on the charge

$q$ is

$\dfrac { q\sigma }{ 2{ \epsilon }_{ 0 } }$ because the electric field due to conducting plate having a charge density

$\sigma$ is

$\dfrac { \sigma }{ 2{ \epsilon }_{ 0 } }$ .

Therefore,

$\tan { \theta } =\dfrac { q\sigma }{ 2{ \epsilon }_{ 0 }mg }$ .

In the diagram, the electric field will be strongest at :

0%
A
0%
B
0%
C
0%
D

Explanation

The intensity of electric field depends upon the density of electric field lines in a region .As electric field lines are denser in the region

$A$ therefore electric field will be strongest in region

$A$ .

An electric filament bulb can be worked from

0%
DC supply only
0%
AC supply only
0%
Battery supply only
0%
All above

Electric charge is developed due to actual transfer of

0%
electron
0%
proton
0%
neutron
0%
none

The safety fuse should have

0%
high resistance and high melting point
0%
high resistance and low melting point
0%
low resistance and high melting point
0%
low resistance and low melting point

The value of

$\dfrac {1}{4\pi \epsilon_{0}}$ is:

0%
$8.854\times 10^{-12}$
0%
$4\pi \times 10^{-7}$
0%
$6.25\times 10^{12}$
0%
$9\times 10^{9}$

Explanation

$\epsilon_o=permittivity\, of\, free \,space$

$\epsilon_o=8.854\times 10^{-12}$

$\dfrac{1}{4\pi\epsilon_o}=9\times 10^9$

Two infinite parallel metal planes, contain electric charges with charge densities

$+\sigma$ and

$-\sigma$ respectively and they are separated by a small distance in air. If the permittivity of air is

$\epsilon_{0}$ , then the magnitude of the field between the two planes with its direction will be:

0%
$\sigma / \epsilon_{0}$ towards the positively charge plane
0%
$\sigma / \epsilon_{0}$ towards the negatively charged plane
0%
$\sigma / (2\epsilon_{0})$ towards the positively charged plane
0%
$0$ and towards any direction

Explanation

Direction of the electric field due to each plate is towards the negatively charged plate and contribution of each plate to the magnitude of field is same, i.e.,

$\dfrac { \sigma }{ 2{ \epsilon }_{ 0 } }$ .

The magnitude of the net field is therefore:

$\dfrac { \sigma }{ { \epsilon }_{ 0 } }$

So, option B is the correct answer.

In the thunderstorm, the charges accumulate near the upper edges of clouds are

0%
negatively charged
0%
positively charged
0%
neutral
0%
none of these

Two uniformly charged plates are pictured above, with four evenly spaced points shown along a horizontal lines between the plates. What is the correct rank of strength of the electric field at the points, greatest first?

0%
1, 2, 3, 4
0%
4, 3, 2, 1
0%
The electric field strength is the same at all points.
0%
1 and 4 tie, 2 and 3 tie
0%
2 and 3 tie, 1 and 4 tie

Explanation

The electric field outsides the plates will be zero and the field in between the plates will be

$E=\dfrac{\sigma}{2\epsilon_0}+\dfrac{\sigma}{2\epsilon_0}=\dfrac{\sigma}{\epsilon_0}$

Thus, electric field in between the plates does not depend on the distance. Hence, the field at every point in between the plates will be same. Thus, option C will correct.

A Gaussian surface in the cylinder of cross section

$\displaystyle \pi { a }^{ 2 }$ and length

$L$ is immersed in a uniform electric field

$\displaystyle \overline { E }$ with the cylinder axis parallel to the field. The flux

$\displaystyle '\phi '$ of the electric field through the closed surface is:

0%
$\displaystyle 2\pi { a }^{ 2 }\overline { E }$
0%
$\displaystyle \pi { a }^{ 2 }\overline { E } L$
0%
$\displaystyle \pi { a }^{ 2 }\left( 2+L \right) \overline { E }$
0%
Zero

Explanation

The electric lines of force would enter from one surface of the cylinder and leave out from the other.

Thus the net flux through the cylinder would be :

$\vec{B}.\vec{A_1}+\vec{B}.\vec{A_2}$

$B(\pi a^2)+B(-\pi a^2)=0$

Two positive charges

$q_1=20C$ and

$q_2=6C$ are separated by diameter of

$3m$ . Then what is the force produced by them?

0%
$80N$
0%
$53.3N$
0%
$56N$
0%
$72.5N$

Explanation

Here, $r=\dfrac{3}{2} =1.5m$ and charges are $q_1=20$ and $q_2=6$

Coulomb's law,

$F=k\dfrac{q_1q_2}{r^2}$

Therefore, (for simplicity assume k=1)

$F=\dfrac{20 \times 6}{(1.5)^2}=53.3N$

4 point charges each +q is placed on the circumference of a circle of diameter 2d in such a way that they form a square. The potential at the centre is:

0%
0
0%
K $\displaystyle \frac { 4q }{ d }$
0%
K $\displaystyle \frac { 4d }{ q }$
0%
K $\displaystyle \frac { q }{ 4d }$

Explanation

Potential at centre due to all charges are

$\displaystyle =\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \left[ \frac { q }{ d } +\frac { q }{ d } +\frac { q }{ d } +\frac { q }{ d } \right]$

$\displaystyle =\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \frac { 4q }{ d }$

$\displaystyle =K\frac { 4q }{ d }$ in CGS units.
1134267_470427_ans_24bd1a2e14d243e2a9bc7c4bd7378787.PNG

A charge

$Q$ , far from other charges, is fixed a distance

$\cfrac{1}{2}s$ above the center of a square with side length s as shown in the diagram. What is the value of the electric flux that passes through the square due to the charge

$Q$ ?

0%
$\cfrac{Q}{{\epsilon}_0}$
0%
$\cfrac{Q}{2{\epsilon}_0}$
0%
$\cfrac{Q}{3{\epsilon}_0}$
0%
$\cfrac{Q}{4{\epsilon}_0}$
0%
$\cfrac{Q}{6{\epsilon}_0}$

Explanation

Let the Gaussian cube of side

$s$ such that the charge is placed at its centre.

Thus electric flux passing through the cube

$\phi_{cube} = \dfrac{Q}{\epsilon_o}$

As the charge is placed symmetric about all the 6 faces of cube, thus flux passing through one face is equal to the one-sixth of the total flux passing through the entire cube.

$\therefore$ Flux passing through the square

$\phi_1 = \dfrac{\phi_{cube}}{6} = \dfrac{Q}{6\epsilon_o}$

If the electric field at all points on a closed Gaussian surface is zero, does this mean that the net charge enclosed by the surface is zero?

0%
Yes; zero field everywhere means zero flux everywhere.
0%
Yes; zero electric field means that there are no charges of any kind in the region.
0%
No; there may be some positive and some negative charges enclosed within the surface.
0%
No; zero field could be caused by charges outside of the surface as well as inside the surface.
0%
No; it is impossible for charges to exist and yet electric field to be zero everywhere.

Explanation

Gauss's theorem states if a closed surface encloses a charge , then surface integral of the electric field over the closed surface is

$1/\varepsilon_{0}$ times of the charge enclosed by the surface ,

$\oint\vec E.\vec {dS}=Q/\varepsilon_{0}$

it is clear that if

$\vec E$ (charge enclosed) is zero then

$Q=0$

Compare how the magnitude of the electric field decreases with distance from a uniform charged sphere and a uniform charged wire.

0%
Both decrease with a one over the square of the distance relationship.
0%
From the sphere, the electric field decreases with a one over the distance relationship. From the wire, the electric field decreases with a one over the distance squared relationship.
0%
Both decrease with a one over the distance relationship.
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From the wire, the electric field decreases with a one over the distance relationship. From the sphere, the electric field decreases with a one over the distance squared relationship.

Explanation

Electric field at a point due to uniform charged wire

$E_w = \dfrac{\lambda}{2\pi \epsilon_o r}$

$\implies E_w =\propto \dfrac{1}{r}$

Thus electric field decreases with one over the distance relationship.

Electric field at a point due to uniform charged sphere

$E_s = \dfrac{Q}{4\pi \epsilon_o r^2}$

$\implies E_s= \propto \dfrac{1}{r^2}$

Thus electric field decreases with one over the distance squared relationship.

The diagram shows three identical metal plates that are set parallel to each other. The left plate is farther from the center plate than the right plate is from the center plate. A source of constant EMF is connected to the system such that the center plate is connected to the positive terminal and the left and right plates are connected to the negative terminal.
Which of the following diagrams correctly depicts the electric field lines (shown as arrows) in the region between the plates?

In the diagram, which of the three charges (1, 2 or 3) is the greatest in magnitude?

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1
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2
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3
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All have the same magnitude

If the net charge enclosed by a closed Gaussian surface is zero, does this mean that the electric field at all points on the surface is also zero?

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Yes; because $\oint \overrightarrow{E}\overrightarrow{dA}=\cfrac{Q_{enc}}{\epsilon_0}$ , if $Q_{enc}$ is zero, then $\overrightarrow {E}$ must be zero.
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Yes; electric field strength depends on how much net charge is present in a region.
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Yes; electric field can be cancelled out by charges outside the surface.
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No; It is not necessary that electric field should be zero but flux should be zero.
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No; there is no relationship between charge and electric field.

Explanation

It is not necessary that electric field should be zero because if net charge is zero. Then there may be possibility that only dot product of

$E$ and

$A$ will be zero. That doesn't mean that electric field will be zero.

An infinite large plate has a net positive charge. When referring to the electric field just as it is coming out of the plate:

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Some of the electric field is perpendicular to the plate
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All of the electric field is perpendicular to the plate
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None of the electric field is perpendicular to the plate
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There is no electric field coming out of the plate

Explanation

From Gauss's law, we find that all the electric field lines coming out of an infinite charged plate are perpendicular to the plate.

Magnitude of electric field

$E = \dfrac{\sigma}{2\epsilon_o}$

The coulombic force of attraction between two charges increases with increase in the distance of separation between them.

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True
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False

Explanation

The coulombic force between two charges is given by

$F=\dfrac{Q_1Q_2}{4\pi\epsilon_o R^2}$

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

Hence on increasing the distance

$R$ , force

$F$ decrease.

Given statement is

$False$ .

Two electric lines of force never intersect.

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True
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False

CBSE Questions for Class 12 Medical Physics Electric Charges And Fields Quiz 5 - 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 5 - MCQExams.com

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Practice Class 12 Medical Physics Quiz Questions and Answers

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