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NEET • PHYSICS • ELECTRIC FIELDS AND CHARGES

Electric fields and charges practice · Hard

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1 Hard

1 of 13

Understanding +4 | -1

Statement-1:

\overrightarrow{{E}}

in outside velocity of a conductor depends only on the local charge density

\alpha

.
Statement-2:

\overrightarrow{{E}}

in outside vicinity of a conductor is given by

\frac{\sigma}{\varepsilon_0}

View answer and explanation

Correct option: D

Statement-1 is false, Statement-2 is true

Option D is correct because Statement-1 is false as the electric field outside a conductor depends on the local charge density, and Statement-2 is true, providing the correct formula.

2 Hard

2 of 13

Application +4 | -1

A charged particle

q_1

is at position

(2,-1,3)

. The electrostatic force on another charge particle

q_2

(0,0,0)

View answer and explanation

Correct option: D

\frac{q_1 q_2}{56 \sqrt{14} \pi \varepsilon_0}(-2 \hat{i}+\hat{j}-3 \hat{k})

Option D is correct because it accurately calculates the electrostatic force vector using Coulomb's law in vector form, considering the position vectors of the charges.

3 Hard

3 of 13

Application +4 | -1

A particle of charge

-q

and mass

m

moves in a circle of radius

r

around an infinitely long line charge of linear charge density

+\lambda.

Then time period will be given as:

(Consider

k

as Coulomb's constant)

View answer and explanation

Correct option: C

T=2 \pi r \sqrt{\frac{m}{2 k \lambda q}}

Option C is correct because it accurately applies the concept of centripetal force provided by the electric field of the line charge to derive the time period of the particle's circular motion.

4 Hard

4 of 13

Application +4 | -1

Two point charges are kept in air with certain separation between them. The force between them is

F_1

, if half of the space between the charges is filled with a medium of dielectric constant 4 and the force between them is

F_2

, if

\frac{1}{3}

rd of the space between the charges is filled with dielectric of dielectric constant 9. Then

F_1 / F_2

View answer and explanation

Correct option: A

\frac{100}{81}

Option A (100/81) is correct because it accurately calculates the ratio of forces with and without the dielectric, considering the effective distance change.

5 Hard

5 of 13

Application +4 | -1

If volume change density inside a solid sphere of radius

{R}

varies with distance

r

from centre of sphere as

\rho=\rho_0 \frac{r}{R} ; \rho_0=

constant then find electric field intensity at distance

{r}(<{R})

from centre is

View answer and explanation

Correct option: A

\frac{\rho_0 r^2}{4 \in_0 R}

Option A (

\frac{\rho_0 r^2}{4 \in_0 R}

) is correct because it is derived from Gauss's law, considering the varying charge density within the sphere.

6 Hard

6 of 13

Application +4 | -1

If the electric flux entering and leaving an enclosed surface respectively is

{\phi _1}

and

{\phi _2}

the electric charge inside the surface will be

View answer and explanation

Correct option: B

\left( {{\phi _2} - {\phi _1}} \right){\varepsilon _0}

Option B ((

{\phi _2} - {\phi _1}

){ε _0}) is correct because Gauss's law relates the net electric flux through a closed surface to the enclosed charge.

7 Hard

7 of 13

Analyzing +4 | -1

Given below are two statements: one is labelled
as Assertion (A) and the other is labelled as
Reason (R).

Assertion (A): Net dipole moment of a polar
linear isotropic dielectric substance is not zero
even in the absence of an external electric field.

Reason (R): In absence of an external electric
field, the different permanent dipoles of a polar
dielectric substance are oriented in random
directions.

In the light of the above statements, choose the
most appropriate answer from the options given
below:

View answer and explanation

Correct option: D

(A) is not correct but (R) is correct

The correct answer is D because the assertion is false; a polar dielectric has zero net dipole moment in the absence of an external field due to random orientation of dipoles. The reason is correct.

8 Hard

8 of 13

Application +4 | -1

A uniformly charged disc of radius

R

having surface charge density

\sigma

is placed in the xy- plane with its centre at the origin. Find the electric field intensity along the

z

–axis at a distance

Z

from origin

View answer and explanation

Correct option: A

E = \frac{\sigma }{{2{\varepsilon _0}}}\left( {1 - \frac{\mathrm{Z}}{{\sqrt {({Z^2} + {R^2})} }}} \right)

Option A (

E = \frac{\sigma }{{2{\varepsilon _0}}}\left( {1 - \frac{\mathrm{Z}}{{\sqrt {({Z^2} + {R^2})} }}} \right)

) is correct because it represents the electric field due to a charged disc along its axis, derived from integrating the contributions of infinitesimal rings.

9 Hard

9 of 13

Understanding +4 | -1

Select the correct statements from the following:

I. The electric field due to a charge outside the Gaussian surface contributes zero net flux through the surface.

II. Total flux linked with a closed body, enclosing any charges will be zero.

III. Total electric flux, if a dipole is enclosed by a surface is zero.

View answer and explanation

Correct option: C

I and III

Option C (I and III) is correct because a charge outside a Gaussian surface contributes zero net flux, and the net flux through a surface enclosing a dipole is zero.

10 Hard

10 of 13

Analyzing +4 | -1

Statement-1: A device used to measure

\overrightarrow{{E}}

is located a some distance from a fixed point charge. In this situation, the device measures

E_0

as the magnitude of electric field intensity. Now an uncharged conducting sphere with a very small hole is lowered by an insulating thread so that it surrounds the point charge. Now, the reading of the device becomes zero.
Statement-2: Electrostatic shielding is the phenomenon in which inside of hollow conductor is shielded for outside electric field.

View answer and explanation

Correct option: C

Statement-1 is true, Statement-2 is false

Statement-1 is true because the electric field inside a conductor is zero. Statement-2 is false because electrostatic shielding is a true phenomenon.

11 Hard

11 of 13

Analyzing +4 | -1

Two equal negative charges

-q

are fixed at the points

(0, a)

and

(0,-a)

on the

y

-axis. A positive charge

Q

is released from rest from the point

(2 a, 0 )

on the

x

-axis. The charge will

View answer and explanation

Correct option: D

Execute oscillatory motion but not S.H.M

Option D is correct because the positive charge will oscillate about the origin, but the restoring force is not proportional to the displacement, hence it's not SHM.

12 Hard

12 of 13

Analyzing +4 | -1

Assertion: On bringing a positively charged rod near the uncharged conductor, the conductor gets attracted towards the rod.
Reason: The electric field on the surface of a conductor is directly proportional to the surface charge density at that point.

View answer and explanation

Correct option: B

If both Assertion and Reason are true, but Reason is not the correct explanation of Assertion

The correct answer is B because both the assertion and reason are true, but the reason does not explain why the conductor is attracted to the rod. The attraction is due to induced charges.

13 Hard

13 of 13

Application +4 | -1

An electric dipole (dipole moment

\mathrm{p}

) is placed at a radial distance

\mathrm{r} \gg \mathrm{a}

(where

\mathrm{a}

is dipole length) from a infinite line of charge having linear charge density

+{\lambda}

. Dipole moment vector is aligned along radial vector

\overrightarrow{\mathrm{r}}

force experienced by dipole is:

View answer and explanation

Correct option: A

\frac{\lambda \mathrm{p}}{2 \pi \varepsilon_0 \mathrm{r}^2}

, attractive

Option A (

\frac{\lambda \mathrm{p}}{2 \pi \varepsilon_0 \mathrm{r}^2}

, attractive) is correct because the force on a dipole in an electric field gradient is proportional to the gradient and the dipole moment, and the force is attractive.

Revision notes and key takeaways

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Key Takeaways

✓ This set contains 13 single-correct-answer MCQs on ELECTRIC FIELDS AND CHARGES.
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