Match List - I with List - II. List - I List - II (A) Electric field inside (distance r > 0 from center) of a uniformly charged spherical shell with surface charge density σ, and radius R. (I) σ/ε0 (B

(A) $\rightarrow 0$ (III) (B) $\rightarrow \frac{\sigma}{2 \varepsilon_0}$ (C) $\rightarrow \frac{\sigma \mathrm{R}^2}{\varepsilon_0 \mathrm{r}^2}$ (No row matching) $(\mathrm{D}) \rightarrow \frac{\sigma}{\varepsilon_0}(\mathrm{I})$

Electrostatics — Page 10 | JEE Physics

Q91

\sigma

is the uniform surface charge density of a thin spherical shell of radius R. The electric field at any point on the surface of the spherical shell is :

A

\sigma / \epsilon_o R

B

\sigma / \in_o

C

\sigma / 2 \epsilon_o

D

\sigma / 4 \epsilon_o

Correct Answer

Option B

Solution

The question is about calculating the electric field at the surface of a thin spherical shell with a uniform surface charge density denoted by $\sigma$ .

To determine the electric field at any point on the surface of the shell, we can use Gauss's law, which is particularly useful for systems with high symmetry like a spherical shell.

Gauss's law in its integral form states that the electric flux through a closed surface is equal to the charge enclosed by the surface divided by the permittivity of free space (

\epsilon_0

), mathematically represented as:

\Phi_E = \frac{Q_{\text{enc}}}{\epsilon_0}

In the case of a spherical shell of radius

R

, the total charge

Q_{\text{enc}}

on the shell can be written in terms of the surface charge density $\sigma$ as:

Q_{\text{enc}} = \sigma \times 4\pi R^2

Now, we apply Gauss's law using a Gaussian surface that coincides with the surface of the spherical shell.

The electric field

E

at the surface of the shell is uniform over the Gaussian surface, and the area of the Gaussian surface (which is also the area of the spherical shell) is

4\pi R^2

. Thus, the electric flux

\Phi_E

through the Gaussian surface is:

\Phi_E = E \cdot 4\pi R^2

Using Gauss's law:

E \cdot 4\pi R^2 = \frac{\sigma \cdot 4\pi R^2}{\epsilon_0}

Simplifying this equation gives us the electric field at the surface of the spherical shell:

E = \frac{\sigma}{\epsilon_0}

Therefore, the correct answer is Option B:

\frac{\sigma}{\epsilon_0}

Q92

Match with .

List - I		List - II
(A)	Electric field inside (distance r > 0 from center) of a uniformly charged spherical shell with surface charge density σ, and radius R.	(I)	σ/ε0
(B)	Electric field at distance r>0 from a uniformly charged infinite plane sheet with surface charge density σ.	(II)	σ/2ε0
(C)	Electric field outside (distance r>0 from center) of a uniformly charged spherical shell with surface charge density σ, and radius R.	(III)	0
(D)	Electric field between 2 oppositely charged infinite plane parallel sheets with uniform surface charge density σ.	(IV)	$\dfrac{\sigma}{\epsilon_0 r^2}$

A (A)-(IV), (B)-(I), (C)-(III), (D)-(II)

B (A)-(II), (B)-(I), (C)-(IV), (D)-(III)

C (A)-(III), (B)-(II), (C)-(IV), (D)-(I)

D (A)-(IV), (B)-(II), (C)-(III), (D)-(I)

Correct Answer

Option C

Solution

(A) $\rightarrow 0$ (III) (B) $\rightarrow \dfrac{\sigma}{2 \varepsilon_0}$ (C) $\rightarrow \dfrac{\sigma \mathrm{R}^2}{\varepsilon_0 \mathrm{r}^2}$ (No row matching) $(\mathrm{D}) \rightarrow \dfrac{\sigma}{\varepsilon_0}(\mathrm{I})$

Q93

The electric flux is

\phi=\alpha \sigma+\beta \lambda

where

\lambda

and

\sigma

are linear and surface charge density, respectively.

\left(\dfrac{\alpha}{\beta}\right)

represents

A displacement

B charge

C electric field

D area

Correct Answer

Option A

Solution

Let's analyze the given expression step by step. We are given the electric flux:

\phi = \alpha \sigma + \beta \lambda

where: $\sigma$ is a surface charge density with units of charge per unit area (C/m²). $\lambda$ is a linear charge density with units of charge per unit length (C/m).

For the equation to be dimensionally consistent, the two terms on the right must have the same units as the flux $\phi$ .

For the term

\alpha \sigma

: Since $\sigma$ has units

\text{C/m}^2

, the constant $\alpha$ must have units that convert $\sigma$ into the same units as the flux. For the term

\beta \lambda

: Since $\lambda$ has units

\text{C/m}

, the constant $\beta$ carries its own units to ensure the term matches the flux’s dimensions.

Since both terms add to give the flux, they must share the same overall dimension.

Thus, the dimensions of

\alpha \sigma

and

\beta \lambda

must be equal:

[\alpha] \cdot \frac{Q}{L^2} = [\beta] \cdot \frac{Q}{L}

Here,

Q

represents the unit of charge and

L

represents a unit of length. Canceling the common factors, we find:

\frac{[\alpha]}{[\beta]} = L

That is, the ratio

\frac{\alpha}{\beta}

has the dimension of length.

Now, let’s match this with the given options: Option A: Displacement (commonly measured as a length) Option B: Charge (has the unit Coulomb) Option C: Electric Field (has units of V/m or N/C) Option D: Area (has units of length squared,

L^2

) Since

\frac{\alpha}{\beta}

has the dimensions of a length, it logically corresponds to a displacement.

Thus, the correct answer is: Option A – displacement.

Q94

Two point charges

-4 \mu \mathrm{c}

and

4 \mu \mathrm{c}

, constituting an electric dipole, are placed at

(-9,0,0) \mathrm{cm}

and

(9,0,0) \mathrm{cm}

in a uniform electric field of strength

10^4 \mathrm{NC}^{-1}

. The work done on the dipole in rotating it from the equilibrium through

180^{\circ}

is :

A 18.4 mJ

B 12.4 mJ

C 16.4 mJ

D 14.4 mJ

Correct Answer

Option D

Solution

The work done on an electric dipole in a uniform electric field when it is rotated from an angle $\theta_1$ to an angle $\theta_2$ is given by:

W = -pE(\cos \theta_2 - \cos \theta_1)

where: $p = q \cdot d$ is the dipole moment, $E$ is the electric field strength, $\theta_1$ and $\theta_2$ are the initial and final angles between the dipole moment and the electric field.

Given: Charge $q = 4 \, \mu \mathrm{C} = 4 \times 10^{-6} \, \mathrm{C}$ Distance $d = 18 \, \mathrm{cm} = 0.18 \, \mathrm{m}$ Electric field $E = 10^4 \, \mathrm{N/C}$ Initial angle $\theta_1 = 0^\circ$ (aligned with the field, equilibrium position) Final angle $\theta_2 = 180^\circ$ First, calculate the dipole moment $p$ :

p = q \cdot d = 4 \times 10^{-6} \, \mathrm{C} \cdot 0.18 \, \mathrm{m} = 7.2 \times 10^{-7} \, \mathrm{Cm}

Calculate the work done $W$ :

W = -pE(\cos 180^\circ - \cos 0^\circ)

W = -7.2 \times 10^{-7} \, \mathrm{Cm} \cdot 10^4 \, \mathrm{N/C} \cdot (-1 - 1)

W = 7.2 \times 10^{-7} \, \mathrm{Cm} \cdot 10^4 \, \mathrm{N/C} \cdot 2

W = 1.44 \times 10^{-2} \, \mathrm{J} = 14.4 \, \mathrm{mJ}

Therefore, the work done on the dipole in rotating it from the equilibrium position through $180^\circ$ is 14.4 mJ (Option D).

Q95

Two charges

7 \mu \mathrm{c}

and

-4 \mu \mathrm{c}

are placed at

(-7 \mathrm{~cm}, 0,0)

and

(7 \mathrm{~cm}, 0,0)

respectively. Given,

\epsilon_0=8.85 \times 10^{-12} \mathrm{C}^2 \mathrm{~N}^{-1} \mathrm{~m}^{-2}

, the electrostatic potential energy of the charge configuration is :

A

-2.0

J

B

-1.5

J

C

-1.2

J

D

-1.8

J

Correct Answer

Option D

Solution

U = \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{r}

Here,

q_1 = 7 \times 10^{-6} \, C

q_2 = -4 \times 10^{-6} \, C

The separation between the charges is along the x-axis from

-7 \, \text{cm}

to

7 \, \text{cm}

, so

r = 0.14 \, m

The Coulomb constant is approximated as

\frac{1}{4\pi\epsilon_0} \approx 9 \times 10^9 \, \frac{Nm^2}{C^2}

Substitute these values into the equation:

U = 9 \times 10^9 \times \frac{(7 \times 10^{-6})(-4 \times 10^{-6})}{0.14}

Calculate the product of the charges:

(7 \times 10^{-6})(-4 \times 10^{-6}) = -28 \times 10^{-12} \, C^2

Now, divide by the distance:

\frac{-28 \times 10^{-12}}{0.14} = -2 \times 10^{-10} \, C^2/m

Finally, multiply by the Coulomb constant:

U = 9 \times 10^9 \times (-2 \times 10^{-10}) = -1.8 \, J

Thus, the electrostatic potential energy of the configuration is:

-1.8 \, J

This corresponds to Option D.

Q96

Consider a parallel plate capacitor of area A (of each plate) and separation '

d

' between the plates. If

E

is the electric field and

\varepsilon_0

is the permittivity of free space between the plates, then potential energy stored in the capacitor is

A

\varepsilon_0 \mathrm{E}^2 \mathrm{Ad}

B

\dfrac{3}{4} \varepsilon_0 \mathrm{E}^2 \mathrm{Ad}

C

\dfrac{1}{4} \varepsilon_0 \mathrm{E}^2 \mathrm{Ad}

D

\dfrac{1}{2} \varepsilon_0 \mathrm{E}^2 \mathrm{Ad}

Correct Answer

Option D

Solution

u = \frac{1}{2}\varepsilon_0 E^2

The energy density of an electric field in free space is given by the expression above.

For a parallel plate capacitor with plate area

A

and separation

d

, the volume between the plates is

V = Ad.

Multiplying the energy density by the volume gives the total energy stored:

U = u \times V = \frac{1}{2}\varepsilon_0 E^2 \times Ad = \frac{1}{2}\varepsilon_0 E^2 Ad.

Thus, the potential energy stored in the capacitor is

\frac{1}{2}\varepsilon_0 E^2 Ad.

This corresponds to the option:

\text{Option D: } \frac{1}{2}\varepsilon_0 E^2 Ad.

Q97

A small uncharged conducting sphere is placed in contact with an identical sphere but having

4 \times 10^{-8} \mathrm{C}

charge and then removed to a distance such that the force of repulsion between them is

9 \times 10^{-3} \mathrm{~N}

. The distance between them is (Take

\dfrac{1}{4 \pi \epsilon_{\mathrm{o}}}

as

9 \times 10^9

in SI units)

A 1 cm

B 2 cm

C 4 cm

D 3 cm

Correct Answer

Option B

Solution

\begin{aligned} & \mathrm{F}=\frac{\mathrm{k}\left(\frac{\theta}{2}\right)\left(\frac{\theta}{2}\right)}{\mathrm{r}^2} \\ & 9 \times 10^{-3}=\frac{9 \times 10^9 \times\left(4 \times 10^{-8}\right) \times 4 \times 10^{-8}}{4 \times \mathrm{r}^2} \\ & \mathrm{r}^2=\frac{9 \times 10^9 \times 16 \times 10^{-16}}{4 \times 9 \times 10^{-3}}=4 \times 10^{-4} \\ & \mathrm{r}=2 \times 10^{-2} \mathrm{~m} \Rightarrow 2 \mathrm{~cm} \end{aligned}

Q98

Three infinitely long wires with linear charge density

\lambda

are placed along the

x-a x i s, y-a x i s

and

z-

axis respectively. Which of the following denotes an equipotential surface?

A

\left(x^2+y^2\right)\left(y^2+z^2\right)\left(z^2+x^2\right)=

constant

B

x y z=

constant

C

x y+y z+z x=

constant

D

(x+y)(y+z)(z+x)=

constant

Correct Answer

Option A

Solution

Electric field due to infinitely long wire

\overrightarrow E = - {{2k\lambda } \over r}\widehat r

We know,

V = - \int {\overrightarrow E \,.\,\overrightarrow {dr} }

= - \int {{{ - 2k\lambda } \over r}\widehat r\,.\,\overrightarrow {dr} }

v = \int {{{2k\lambda } \over r}dr = 2k\lambda \,\ln r + c}

Net potential due to all wires,

V = {v_1} + {v_2} + {v_3}

= 2k\lambda \ln \left( {\sqrt {{x^2} + {y^2}} } \right) + 2k\lambda \ln \left( {\sqrt {{y^2} + {z^2}} } \right) + 2k\lambda \ln \left( {\sqrt {{z^2} + {x^2}} } \right) + c

for equipotential surface,

v = c

\Rightarrow 2k\lambda \left[ {\ln \left( {\sqrt {{x^2} + {y^2}} \,.\,\sqrt {{y^2} + {z^2}} \,.\,\sqrt {{z^2} + {x^2}} } \right)} \right] = c

\Rightarrow \sqrt {({x^2} + {y^2})\,.\,({y^2} + {z^2})\,.\,({z^2} + {x^2})} = c

\Rightarrow ({x^2} + {y^2})({y^2} + {z^2})({z^2} + {x^2}) = c

where, c = constant.

Q99

Two metal spheres of radius R and 3R have same surface charge density σ. If they are brought in contact and then separated, the surface charge density on smaller and bigger sphere becomes σ1 and σ2, respectively. The ratio

\dfrac{\sigma_1}{\sigma_2}

is

A

\dfrac{1}{3}

B

\dfrac{1}{9}

C 9

D 3

Correct Answer

Option D

Solution

When two conducting spheres are in contact, they achieve the same electric potential.

Consider two metal spheres with radii $R$ and $3R$ that have the same surface charge density $\sigma$ .

For a conductive sphere, the potential $V$ of a sphere can be expressed as $V = \dfrac{\sigma r}{\varepsilon_0}$ , where $r$ is the radius of the sphere and $\varepsilon_0$ is the permittivity of free space.

When the two spheres are brought into contact and then separated, they equalize their potentials: $V_1 = V_2$ Thus, the equations become: $\sigma_1 r_1 = \sigma_2 r_2$ Substituting for the radii: $\sigma_1 R = \sigma_2 (3R)$ From this, we can solve for the ratio of the surface charge densities: $\dfrac{\sigma_1}{\sigma_2} = \dfrac{3R}{R} = 3$ Therefore, the ratio of the surface charge densities on the smaller sphere to the larger sphere after contact is 3.

Q100

Given below are two statements: one is labelled as Assertion

\mathbf{A}

and the other is labelled as Reason

\mathbf{R}

Assertion A : Work done in moving a test charge between two points inside a uniformly charged spherical shell is zero, no matter which path is chosen. Reason R : Electrostatic potential inside a uniformly charged spherical shell is constant and is same as that on the surface of the shell. In the light of the above statements, choose the correct answer from the options given below.

A A is true but R is false

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

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

D A is false but R is true

Correct Answer

Option B

Solution

\textbf{Assertion A:}

For a uniformly charged spherical shell, Gauss's law tells us that the electric field inside the shell is zero.

Since the electric field

\vec{E}

is zero everywhere inside, the potential

V

must be constant.

Work done in moving a charge from one point to another in an electric field is given by the difference in potential energy, which is

q(V_B - V_A)

. If the potential difference

V_B - V_A

is zero (because the potential is constant), then the work done is zero, regardless of the path taken.

Therefore, Assertion A is true.

\textbf{Reason R:}

It states that the electrostatic potential inside a uniformly charged spherical shell is constant and equal to that on its surface.

This is correct because the electric field inside is zero, ensuring that the potential remains uniform inside.

Thus, Reason R is also true.

\textbf{Relationship between A and R:}

The work done on a test charge moving inside the shell being zero is a direct consequence of the fact that the potential is constant inside.

Hence, Reason R correctly explains why the work done (Assertion A) is zero.

Given this analysis, the correct answer is: Option B "Both A and R are true and R is the correct explanation of A."