HARD

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A non-conducting sphere of radius $R$ has a uniform volume charge density $ρ$ . A spherical cavity of radius $b$ is created inside the sphere, whose centre lies at a distance $a$ from the centre of sphere. It can be shown that field inside cavity will be uniform

What is electric field at any point inside the spherical cavity?

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Important Questions on Electrostatics

HARD

Physics>Electricity and Magnetism>Electrostatics>Gauss’s Law

The potential (in volts) of a charge distribution is given by

V (z) = 30 - 5 z^{2}

for

|z| \leq 1 m

V (z) = 35 - 10 |z|

for

|z| \geq 1 m

V (z)

does not depend on x and y. If this potential is generated by a constant charge per unit volume

ρ_{0}

(in units of

ϵ_{0}

) which is spread over a certain region, then choose the correct statement.

EASY

Physics>Electricity and Magnetism>Electrostatics>Gauss’s Law

A charge

+ q

is at a distance

\frac{L}{2}

above a square of side

L

. Then what is the flux linked with the surface?

HARD

Physics>Electricity and Magnetism>Electrostatics>Gauss’s Law

The region between two concentric spheres of radii 'a' and 'b', respectively (see figure), has volume charge density

ρ = \frac{A}{r}

, where A is a constant and r is the distance from the centre. At the centre of the spheres is a point charge Q. The value of A such that the electric field in the region between the spheres will be constant, is:

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MEDIUM

Physics>Electricity and Magnetism>Electrostatics>Gauss’s Law

A conducting sphere of radius

R

is given a charge

Q .

The electric potential and the electric field at the center of the sphere respectively are:

EASY

Physics>Electricity and Magnetism>Electrostatics>Gauss’s Law

If some charge is given to a solid metallic sphere, the field inside remains zero and by Gauss's law all the charge resides on the surface. Suppose now that Colomb's force between two charges varies as

\frac{1}{r^{3}} .

Then, for a charged solid metallic sphere

EASY

Physics>Electricity and Magnetism>Electrostatics>Gauss’s Law

A hollow metal sphere of radius

R

is uniformly charged. The electric field due to the sphere at a distance

r

from the center

EASY

Physics>Electricity and Magnetism>Electrostatics>Gauss’s Law

The electric field in a region of space is given by,

\vec{E} = E_{0} \hat{i} + 2 E_{0} \hat{j}

where

E_{0} = 100 N C^{- 1}

. The flux of this field through a circular surface of radius

0.02 m

parallel to the

Y ‐ Z

plane is nearly

HARD

Physics>Electricity and Magnetism>Electrostatics>Gauss’s Law

A circular disc of radius

R

carries surface charge density

σ (r) = σ_{0} (1 - \frac{r}{R})

, where

σ_{0}

is a constant and

r

is the distance from the center of the disc. Electric flux through a large spherical surface that encloses the charged disc completely is

ϕ_{0}

. Electric flux through another spherical surface of radius

\frac{R}{4}

and concentric with the disc is

ϕ

. Then the ratio

\frac{ϕ_{0}}{ϕ}

EASY

Physics>Electricity and Magnetism>Electrostatics>Gauss’s Law

A charged particle moves with a velocity

v

in a circular path of radius

R

around a long uniformly charged conductor, then

EASY

Physics>Electricity and Magnetism>Electrostatics>Gauss’s Law

Four closed surfaces and corresponding charge distributions are shown below.

Question Image

Let the respective electric fluxes through the surfaces be

ϕ_{1}, ϕ_{2}, ϕ_{3}

and

ϕ_{4}

. Then:

HARD

Physics>Electricity and Magnetism>Electrostatics>Gauss’s Law

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?

EASY

Physics>Electricity and Magnetism>Electrostatics>Gauss’s Law

A hollow insulated conducting sphere is given a positive charge of

10 μC

. What will be the electric field at the centre of the sphere? The radius of the sphere is

2 m

EASY

Physics>Electricity and Magnetism>Electrostatics>Gauss’s Law

The electric flux (in

SI

units) through any face of a cube due to a positive charge

Q

situated at the centre of a cube is

MEDIUM

Physics>Electricity and Magnetism>Electrostatics>Gauss’s Law

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

EASY

Physics>Electricity and Magnetism>Electrostatics>Gauss’s Law

Consider the force

F

on a charge '

q

' due to a uniformly charged spherical shell of radius

R

carrying charge

Q

distributed uniformly over it. Which one of the following statements is true for

F

, if '

q

' is placed at distance

r

from the centre of the shell?

EASY

Physics>Electricity and Magnetism>Electrostatics>Gauss’s Law

A point charge $q$ is placed at the corner of a cube of side $a$ as shown in the figure. What is the electric flux through the face $A B C D$ ?

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MEDIUM

Physics>Electricity and Magnetism>Electrostatics>Gauss’s Law

In the following figures, which figure represents the variation of the electric field with distance from the centre of a uniformly charged non-conducting spheres of radius

R

EASY

Physics>Electricity and Magnetism>Electrostatics>Gauss’s Law

The magnitude of the average electric field normally present in the atmosphere just above the surface of the Earth is about

150 N / C

, directed inward towards the center of the Earth. This gives the total net surface charge carried by the Earth to be : [Given :

\in_{O} = 8.85 \times 1 0^{- 12} C^{2} / {N-m}^{2}, R_{E} = 6.37 \times 1 0^{6} m

]

EASY

Physics>Electricity and Magnetism>Electrostatics>Gauss’s Law

Shown in the figure are two point charges

+ Q

and

- Q

inside the cavity of a spherical shell. The charges are kept near the surface of the cavity on opposite sides of the centre of the shell. If

σ_{1}

is the surface charge on the inner surface and

Q_{1}

net charge on it and

σ_{2}

the surface charge on the outer surface and

Q_{2}

net charge on it then:

Question Image

EASY

Physics>Electricity and Magnetism>Electrostatics>Gauss’s Law

A charge

Q

is placed at a distance

\frac{a}{2}

above the centre of a square surface of side length

a

. The electric flux through the square surface due to the charge would be?
Question Image

Important Questions on Electrostatics

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