If ∮ S E → · d S → = 0 over a surface, then:

The number of flux lines entering the surface must be equal to the number of flux lines leaving it.

The electric field inside the surface is necessarily uniform.

The magnitude of electric field on the surface is constant.

All the charges must necessarily be inside the surface.

EASY

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The magnitude of electric flux through area enclosed in a square frame $L M N O$ is,

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Important Questions on Electric Charges and Fields

HARD

Physics>Electricity and Magnetism>Electric Charges and Fields>Electric Flux

An infinitely long thin non-conducting wire is parallel to the z - axis and carries a uniform line charge density

λ

. It pierces a thin non-conducting spherical shell of radius

R

in such a way that the arc

P Q

subtends an angle

120^{o}

at the centre

O

of the spherical shell, as shown in the figure. The permittivity of free space is

ϵ_{0}

. Which of the following statements is (are) true?
Question Image

HARD

Physics>Electricity and Magnetism>Electric Charges and Fields>Electric Flux

The electric field in a region is given by

\vec{E} = (\frac{3}{5} E_{0} \hat{i} + \frac{4}{5} E_{0} \hat{j}) N C^{- 1}

. The ratio of flux of reported field through the rectangular surface of area

0.2 m^{2}

(parallel to

y - z

plane) to that of the surface of area

0.3 m^{2}

(parallel to

x - z

plane) is

a : b = a : 2,

where

a = ?

[Here

\hat{i}, \hat{j}

and

\hat{k}

are unit vectors along

x, y

and

z

-axes respectively]

EASY

Physics>Electricity and Magnetism>Electric Charges and Fields>Electric Flux

\oint_{S} \vec{E} \cdot d \vec{S} = 0

over a surface, then:

HARD

Physics>Electricity and Magnetism>Electric Charges and Fields>Electric Flux

Two non-conducting spheres of radii $R_{1}$ and $R_{2}$ and carrying uniform volume charge densities $+ ρ$ and $- ρ$ , respectively, are placed such that they partially overlap, as shown in the figure. At all points in the overlapping region,
Question Image

HARD

Physics>Electricity and Magnetism>Electric Charges and Fields>Electric Flux

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>Electric Charges and Fields>Electric Flux

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>Electric Charges and Fields>Electric Flux

The electric field in a region is given by

\vec{E} = \frac{2}{5} E_{0} \hat{i} + \frac{3}{5} E_{0} \hat{j}

with

E_{0} = 4.0 \times 10^{3} N C^{- 1} .

The flux of this field through a rectangular surface, area

0.4 m^{2}

parallel to the

Y - Z

plane is _______

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

HARD

Physics>Electricity and Magnetism>Electric Charges and Fields>Electric Flux

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>Electric Charges and Fields>Electric Flux

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 $A B C D$ ?

Question Image

EASY

Physics>Electricity and Magnetism>Electric Charges and Fields>Electric Flux

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

EASY

Physics>Electricity and Magnetism>Electric Charges and Fields>Electric Flux

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:

EASY

Physics>Electricity and Magnetism>Electric Charges and Fields>Electric Flux

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>Electric Charges and Fields>Electric Flux

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:

Question Image

EASY

Physics>Electricity and Magnetism>Electric Charges and Fields>Electric Flux

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

HARD

Physics>Electricity and Magnetism>Electric Charges and Fields>Electric Flux

Consider a uniform spherical charge distribution of radius

R_{1}

centred at the origin O. In this distribution, a spherical cavity of radius

R_{2}

, centred at

P

with distance

O P = a = R_{1} - R_{2}

(see figure) is made. If the electric field inside the cavity at position

\vec{r}

\vec{E} (\vec{r})

, then the correct statement(s) is (are)

Question Image

HARD

Physics>Electricity and Magnetism>Electric Charges and Fields>Electric Flux

A charged shell of radius

R

carries a total charge

Q .

Given

Φ

as the flux of electric field through a closed cylindrical surface of height

h,

radius

r

and with its center same as that of the shell. Here, center of the cylinder is a point on the axis of the cylinder which is equidistant from its top and bottom surfaces. Which of the following option(s) is/are correct? [

\in_{0}

is the permittivity of free space]

EASY

Physics>Electricity and Magnetism>Electric Charges and Fields>Electric Flux

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>Electric Charges and Fields>Electric Flux

A charge $q$ is placed at one corner of a cube as shown in figure. The flux of electrostatic field $\vec{E}$ through the shaded area is:

Question Image

MEDIUM

Physics>Electricity and Magnetism>Electric Charges and Fields>Electric Flux

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>Electric Charges and Fields>Electric Flux

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

]

The magnitude of electric flux through area enclosed in a square frame LMNO is,

Important Questions on Electric Charges and Fields

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The magnitude of electric flux through area enclosed in a square frame $L M N O$ is,