Given below are two statements : One is labelled as Assertion A and the other is labelled as Reason R. Assertion A : Two metallic spheres are charged to the same potential. One of them is hollow and another is solid, and both have the same radii. Solid sphere will have lower charge than the hollow one. Reason R : Capacitance of metallic spheres depend on the radii of spheres. In the light of the above statements, choose the correct answer from the options given below.

Both A and R are true and R is the correct explanation of A

Both A and R are true but R is not the correct explanation of A

Let σ be the uniform surface charge density of two infinite thin plane sheets shown in figure. Then the electric fields in three different region E I , E I I and E I I I

E → I = - σ ∈ 0 n ^ , E → I I = 0 , E → I I I = σ ∈ 0 n ^

E → I = 2 σ ∈ 0 n ^ , E → I I = 0 , E → I I I = 2 σ ∈ 0 n ^

E → I = 0 , E → I I = σ ∈ 0 n ^ , E → I I I = 0

E → I = σ 2 ∈ 0 n ^ , E → I I = 0 , E → I I I = σ 2 ∈ 0 n ^

Considering a group of positive charges, which of the following statements is correct?

Net potential of the system cannot be zero at a point but net electric field can be zero at that point

Net potential of the system at a point can be zero but net electric field can't be zero at that point

Both the net potential and the net field can be zero at a point

Both the net potential and the net electric field cannot be zero at a point

HARD

JEE Main

IMPORTANT

Earn 100

In finding the electric field using Gauss law the formula $|\vec{E}| = \frac{q_{e n c}}{ε_{0} |A|}$ is applicable. In the formula $ε_{0}$ is permittivity of free space, $A$ is the area of Gaussian surface and $q_{e n c}$ is charge enclosed by the Gaussian surface. This equation can be used in which of the following situation?

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Important Questions on Electrostatics

HARD

JEE Main

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR PHYSICS>Chapter 13 - Electrostatics>JEE Main - 8th January 2020 Shift 2>Q 81

Consider two charged metallic spheres

S_{1}

and

S_{2}

of radii

R_{1}

and

R_{2},

respectively. The electric fields

E_{1}

(on

S_{1}

) and

E_{2}

(on

S_{2}

) on their surfaces are such that

\frac{E_{1}}{E_{2}} = \frac{R_{1}}{R_{2}} .

Then the ratio

V_{1}

(on

S_{1}

)

/

V_{2}

(on

S_{2}

) of the electrostatic potentials on each sphere is:

HARD

JEE Main

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR PHYSICS>Chapter 13 - Electrostatics>JEE Main - 8th January 2020 Shift 2>Q 82

A capacitor is made of two square plates each of side

a

making a very small angle

α

between them, as shown in figure. The capacitance will be close to:
Question Image

HARD

JEE Main

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR PHYSICS>Chapter 13 - Electrostatics>JEE Main - 8th January 2020 Shift 2>Q 83

A particle of mass

m

and charge

q

is released from rest in a uniform electric field. If there is no other force on the particle, the dependence of its speed

v

on the distance

x

travelled by it is correctly given by (graphs are schematic and not drawn to scale)

HARD

JEE Main

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR PHYSICS>Chapter 13 - Electrostatics>JEE Main - 9th January 2020 Shift 1>Q 84

Consider a sphere of radius

R

which carries a uniform charge density

ρ

. If a sphere of radius

\frac{R}{2}

is carved out of it, as shown, the ratio

\frac{|\vec{E_{A}}|}{|\vec{E_{B}}|}

of magnitude of electric field

\vec{E_{A}}

and

\vec{E_{B}}

, respectively, at points

A

and

B

due to the remaining portion is:

Question Image

MEDIUM

JEE Main

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR PHYSICS>Chapter 13 - Electrostatics>JEE Main - 9th January 2020 Shift 1>Q 85

An electric dipole of moment

\vec{p} = (- \hat{i} - 3 \hat{j} + 2 \hat{k}) \times 10^{- 29} C m

at the origin

(0, 0, 0)

. The electric field due to this dipole at

\vec{r} = + \hat{i} + 3 \hat{j} + 5 \hat{k}

(note that

\vec{r} \cdot \vec{p} = 0

) is parallel to:

HARD

JEE Main

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR PHYSICS>Chapter 13 - Electrostatics>JEE Main - 9th January 2020 Shift 1>Q 86

A charged particle of mass

m

and charge

q

moving under the influence of uniform electric field

\vec{E} \hat{i}

and a uniform magnetic field

\vec{B} \hat{k}

follows a trajectory from point

P

Q

as shown in figure. The velocities at

P

and

Q

are respectively,

v \vec{i}

and

- 2 v \vec{j}

. Then which of the following statements (A, B, C, D) are the correct? (Trajectory shown is schematic and not to scale)

Question Image

(A)

E = \frac{3}{4} (\frac{m v^{2}}{q a})

(B)

Rate of work done by the electric field at P is

\frac{3}{4} (\frac{m v^{3}}{a})

(C)

Rate of work done by both the fields at Q is zero

(D)

The difference between the magnitude of angular momentum of the particle at P and Q is

2 m a v

HARD

JEE Main

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR PHYSICS>Chapter 13 - Electrostatics>JEE Main - 9th January 2020 Shift 2>Q 87

Two identical capacitors

A

and

B,

charged to the same potential

5 V

are connected in two different circuits as shown below at time

t = 0 .

If the charge on capacitors

A

and

B

at time

t = C R

Q_{A}

and

Q_{B}

respectively, then (Here

e

is the base of natural logarithm)

Question Image

MEDIUM

JEE Main

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR PHYSICS>Chapter 13 - Electrostatics>JEE Main - 9th January 2020 Shift 2>Q 88

An electric field

\vec{E} = [4 x \hat{i} - (y^{2} + 1) \hat{j}] N C^{- 1}

passes through the box shown in figure. The flux of the electric field through surfaces

A B C D

and

B C G F

are marked as

ϕ_{I}

and

ϕ_{I I}

respectively. The value of

|ϕ_{I} - ϕ_{I I}|

is (in

N m^{2} C^{- 1}

) ____________.
Question Image

In finding the electric field using Gauss law the formula E→=qencε0A is applicable. In the formula ε0 is permittivity of free space, A is the area of Gaussian surface and qenc is charge enclosed by the Gaussian surface. This equation can be used in which of the following situation?

Important Questions on Electrostatics

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