Electrostatics — Page 11 | JEE Physics

Q101

An infinitely long wire has uniform linear charge density

\lambda = 2 \text{ nC/m}

. The net flux through a Gaussian cube of side length

\sqrt{3}

cm, if the wire passes through any two corners of the cube, that are maximally displaced from each other, would be

x \text{ Nm}^2\text{C}^{-1}

, where

x

is: [Neglect any edge effects and use

\dfrac{1}{4\pi \epsilon_0} = 9 \times 10^9

SI units]

A

6.48 \pi

B

0.72 \pi

C

1.44 \pi

D

2.16 \pi

Correct Answer

Option D

Solution

To find the net flux through a Gaussian cube when an infinitely long wire with a uniform linear charge density $\lambda = 2 \text{ nC/m}$ passes through the cube's diagonal corners, we use Gauss's Law.

The relevant expression is: $\phi = \dfrac{q_{\text{enc}}}{\varepsilon_0}$ Here, $q_{\text{enc}}$ is the charge enclosed by the Gaussian surface, and $\varepsilon_0$ is the permittivity of free space.

Given: The wire passes through two opposite corners of the cube.

The side length of the cube, $a = \sqrt{3} \text{ cm} = \sqrt{3} \times 10^{-2} \text{ m}$ .

The length of the wire inside the cube is equal to its diagonal, which is $\sqrt{3} \times a$ .

Thus, the charge enclosed $q_{\text{enc}}$ is: $q_{\text{enc}} = \lambda \times \text{(length of wire inside the cube)} = \lambda \times \sqrt{3} \times a$ Substitute the given values: $\lambda = 2 \times 10^{-9} \text{ C/m}, \quad a = \sqrt{3} \times 10^{-2} \text{ m}$ Calculating: $q_{\text{enc}} = 2 \times 10^{-9} \times \sqrt{3} \times \sqrt{3} \times 10^{-2}$ Now apply Gauss's Law for net flux: $\phi = \dfrac{\lambda \cdot \sqrt{3} \cdot a}{\varepsilon_0}$ Substitute $\varepsilon_0 = \dfrac{1}{4\pi \times 9 \times 10^9}$ : $\phi = 2 \times 10^{-9} \times \sqrt{3} \times \sqrt{3} \times 10^{-2} \times 36 \pi \times 10^9$ $\phi = 2.16 \pi \text{ Nm}^2\text{C}^{-1}$ Thus, the net flux through the Gaussian cube is $2.16 \pi \text{ Nm}^2\text{C}^{-1}$ .

Q102

A point charge

+q

is placed at the origin. A second point charge

+9 q

is placed at (

\mathrm{d}, 0,0

) in Cartesian coordinate system. The point in between them where the electric field vanishes is:

A

(3 \mathrm{d} / 4,0,0)

B

(\mathrm{d} / 4,0,0)

C

(4 \mathrm{d} / 3,0,0)

D

(\mathrm{d} / 3,0,0)

Correct Answer

Option B

Solution

Let $E_p=0$

\begin{aligned} & \therefore \frac{k q}{x^2}=\frac{k 9 q}{(d-x)^2} \\ & \Rightarrow \frac{d-x}{x}=3 \Rightarrow x=\frac{d}{4} \end{aligned}

$\therefore$ co-ordinate of P is $\left(\dfrac{\mathrm{d}}{4}, 0,0\right)$

Q103

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 :

A Both (A) and (R) are correct and (R) is the correct explanation of (A) 2.

B Both (A) and (R) are correct but (R) is not the correct explanation of (A)

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

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

Correct Answer

Option C

Solution

Assertion (A): The net dipole moment of a polar linear isotropic dielectric substance is not zero, even when there is no external electric field.

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

Explanation: Assertion (A): The statement is incorrect.

In a polar linear isotropic dielectric substance, the net dipole moment is zero when an external electric field is absent.

This is because the dipoles within the material are randomly oriented and cancel each other out.

Reason (R): The statement is correct.

In the absence of an external electric field, the dipoles in a polar dielectric substance remain inherently polarized but are oriented randomly.

This randomness in orientation leads to a cancellation of the net dipole moment.

Therefore, while the reason is correct, it does not support the validity of the assertion as stated.

Q104

Consider a circular loop that is uniformly charged and has a radius

\mathrm{a} \sqrt{2}

. Find the position along the positive

z

-axis of the cartesian coordinate system where the electric field is maximum if the ring was assumed to be placed in

x y

plane at the origin :

A a

B

a / 2

C 0

D

a / \sqrt{2}

Correct Answer

Option A

Solution

To determine the position along the positive $z$ -axis where the electric field due to a uniformly charged circular loop is at its maximum, we follow these steps: Expression for Electric Field ( $E$ ): The electric field $E$ at a point along the $z$ -axis for a charged circular loop can be expressed as: $E = \dfrac{KQr}{(x^2 + R^2)^{3/2}}$ Here, $K$ is the Coulomb's constant, $Q$ is the total charge, $r$ is a constant involving charge distribution, $x$ is the distance along the $z$ -axis, and $R$ is the radius of the loop.

Maximizing the Electric Field: To find where $E$ is maximum, we take the derivative of $E$ with respect to $x$ and set it to zero: $\dfrac{dE}{dx} = 0$ Solve for $x$ : Solving the equation from the derivative, we find: $x = \dfrac{R}{\sqrt{2}}$ Substitute Given Radius: Given the radius $R = a\sqrt{2}$ , substituting into the expression for $x$ : $x = \dfrac{\sqrt{2}a}{\sqrt{2}} = a$ Thus, the position along the positive $z$ -axis where the electric field is maximum is at $x = a$ .

Q105

A dipole with two electric charges of 2 µC magnitude each, with separation distance 0.5 µm, is placed between the plates of a capacitor such that its axis is parallel to an electric field established between the plates when a potential difference of 5 V is applied. Separation between the plates is 0.5 mm. If the dipole is rotated by 30° from the axis, it tends to realign in the direction due to a torque. The value of torque is:

A 2.5×10−9 Nm

B 2.5×10−12 Nm

C 5×10−3 Nm

D 5×10−9 Nm

Correct Answer

Option D

Solution

Given: Potential difference $V = 5 \, \text{V}$ Separation between the plates $d = 0.5 \, \text{mm} = 0.5 \times 10^{-3} \, \text{m}$ The electric field $E$ is calculated as: $E = \dfrac{V}{d} = \dfrac{5}{0.5 \times 10^{-3}} = 10^4 \, \text{V/m}$ The torque $\tau$ experienced by the dipole in the electric field is given by: $\tau = PE \sin \theta$ Where: $P$ is the dipole moment, $E$ is the electric field, $\theta = 30^\circ$ is the angle of rotation.

The dipole moment $P$ is calculated by: $P = q \times a = (2 \times 10^{-6} \, \text{C}) \times (0.5 \times 10^{-6} \, \text{m}) = 1 \times 10^{-12} \, \text{C} \cdot \text{m}$ Substituting the values into the torque equation: $\tau = (1 \times 10^{-12} \, \text{C} \cdot \text{m}) \times 10^4 \, \text{V/m} \times \sin 30^\circ$ Since $\sin 30^\circ = 0.5$ , the torque simplifies to: $\tau = 1 \times 10^{-12} \times 10^4 \times 0.5 = 5 \times 10^{-9} \, \text{N} \cdot \text{m}$ Thus, the torque is $5 \times 10^{-9} \, \text{N} \cdot \text{m}$ .

Q106

Given below are two statements : one is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A) : The outer body of an aircraft is made of metal which protects persons sitting inside from lightning strikes. Reason (R) : The electric field inside the cavity enclosed by a conductor is zero. In the light of the above statements, choose the most appropriate answer from the options given below :

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

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

C Both (A) and (R) are correct and (R) is the correct explanation of (A)

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

Correct Answer

Option C

Solution

The assertion suggests that the outer body of an aircraft is made of metal to protect passengers from lightning strikes.

The reason provided is that the electric field inside a cavity enclosed by a conductor is zero.

This reasoning is based on the property of metallic conductors, where the electric field within a conductor's cavity is indeed zero.

This phenomenon occurs because the charges on a conductor redistribute themselves to cancel any external electric fields within the conductor.

Thus, if lightning were to strike an aircraft, the metal body would direct the electric charges along its surface, effectively shielding the interior where passengers are located.

Therefore, the assertion and the reason are both correct, and the reason correctly explains why the metal body of an aircraft shields passengers from lightning.

This aligns with the principles of electromagnetic shielding.

The key point is that the conductive nature of the aircraft’s metal body ensures no electric field permeates the interior, keeping passengers safe.

Q107

The electrostatic potential on the surface of uniformly charged spherical shell of radius

\mathrm{R}=10 \mathrm{~cm}

is 120 V . The potential at the centre of shell, at a distance

\mathrm{r}=5 \mathrm{~cm}

from centre, and at a distance

\mathrm{r}=15

cm from the centre of the shell respectively, are:

A

0 \mathrm{~V}, 120 \mathrm{~V}, 40 \mathrm{~V}

B

120 \mathrm{~V}, 120 \mathrm{~V}, 80 \mathrm{~V}

C

40 \mathrm{~V}, 40 \mathrm{~V}, 80 \mathrm{~V}

D

0 \mathrm{~V}, 0 \mathrm{~V}, 80 \mathrm{~V}

Correct Answer

Option B

Solution

The potential inside a uniformly charged spherical shell is equal to the potential on its surface.

This means: $V_{\text{in}} = V_{\text{surface}} = \dfrac{kQ}{R} = 120 \, \text{V}$ This maintains the potential at both the center of the shell and any point inside it up to the surface.

Therefore, the potential at the center of the shell and at any point inside the shell (like at a distance $r = 5 \, \text{cm}$ ) is also $120 \, \text{V}$ .

For a point outside the shell, at a distance $r = 15 \, \text{cm}$ , the potential is given by the formula: $V = \dfrac{kQ}{r} = \dfrac{120 \times 10}{15} = 80 \, \text{V}$

Q108

Two small spherical balls of mass 10 g each with charges

-2 \mu \mathrm{C}

and

2 \mu \mathrm{C}

, are attached to two ends of very light rigid rod of length 20 cm . The arrangement is now placed near an infinite nonconducting charge sheet with uniform charge density of

100 \mu \mathrm{C} / \mathrm{m}^2

such that length of rod makes an angle of

30^{\circ}

with electric field generated by charge sheet. Net torque acting on the rod is: (Take

\varepsilon_{\mathrm{o}}: 8.85 \times 10^{-12} \mathrm{C}^2 / \mathrm{Nm}^2

)

A 1.12 Nm

B 2.24 Nm

C 11.2 Nm

D 112 Nm

Correct Answer

Option A

Solution

\begin{aligned} & \mathrm{E}=\frac{\sigma}{2 \varepsilon_0} \\ & \tau=\mathrm{PE} \sin \theta \\ & =\left[\left(2 \times 10^{-6}\right)\left(\frac{2}{10}\right)\right]\left[\frac{100 \times 10^{-6}}{2 \times 8.85 \times 10^{-12}}\right]\left(\frac{1}{2}\right) \\ & =\frac{10}{8.85}=1.12 \mathrm{Nm} \end{aligned}

Q109

Two point charges q1

\left( {\sqrt {10} \mu C} \right)

and q2(

-

25

\mu

C) are placed on the x-axis at x = 1 m and x = 4 m respectively. The electric field (in V/m) at a point y = 3 m on y-axis is, [take

{1 \over {4\pi { \in _0}}}

= 9

\times

109 Nm2C

-

2]

A

\left( {63\widehat i - 27\widehat j} \right) \times {10^2}

B

\left( { - 63\widehat i + 27\widehat j} \right) \times {10^2}

C

\left( {81\widehat i - 81\widehat j} \right) \times {10^2}

D

\left( { - 81\widehat i + 81\widehat j} \right) \times {10^2}

Correct Answer

Option A

Solution

Electric field due to

\sqrt {10} \,\mu C

charge :

\overrightarrow {{E_1}} = -

E1 sin $\theta$ 1

\widehat i

+ E1 cos $\theta$ 1

\widehat j

Where, E1

= {1 \over {4\pi {\varepsilon _0}}} \times {{\left| {{q_1}} \right|} \over {{r_1}^2}}

= 9 \times {10^9} \times {{\sqrt {10} \times {{10}^{ - 6}}} \over {{{\left( {\sqrt {{1^2}} + {3^2}} \right)}^2}}}

= {{9 \times {{10}^3}} \over {\sqrt {10} }}\,v/m

sin $\theta$ 1 $=$

{1 \over {\sqrt {10} }}

and cos $\theta$ 1 =

{3 \over {\sqrt {10} }}

$\therefore$

\overrightarrow {{E_1}} = {{9 \times {{10}^3}} \over {\sqrt {10} }}\,\left( { - {1 \over {10}}\widehat i + {3 \over {\sqrt {10} }}\widehat j} \right)

= 9 \times {10^2}\left( { - \widehat i + 3\widehat j} \right)

Electric field due to $-$ 25 $\mu$ C charge,

\overrightarrow {{E_2}} =

E2 sin $\theta$ 2

\widehat i

$-$ E2 cos $\theta$ 2

\widehat j

where E2

= {1 \over {4\pi {\varepsilon _0}}} \times {{\left| {{9_2}} \right|} \over {r_2^2}}

= 9 \times {10^9} \times {{25 \times {{10}^{ - 6}}} \over {{{\left( {\sqrt {{4^2} + {3^2}} } \right)}^2}}}

= 9 \times {10^3}

V/m sin $\theta$ 2 =

{4 \over 5}

and cos $\theta$ 2 =

{3 \over 5}

$\therefore$

\overrightarrow {{E_2}} = 9 \times {10^3}\,\,\left( {{4 \over 5}\widehat i - {3 \over 5}\widehat j} \right)

= 18 \times {10^2}\left( {4\widehat i - 3\widehat j} \right)

$\therefore$ Net electric field,

\overrightarrow E

=

{\overrightarrow E _1}

+

{\overrightarrow E _2}

= \left( {63\widehat i - 27\widehat j} \right) \times {10^2}\,\,V/m

Q110

Three charges + Q, q, + Q are placed respectively, at distance, 0, d/2 and d from the origin, on the x-axis. If the net force experienced by + Q, placed at x = 0, is zero, then value of q is :

A

-

{Q \over 4}

B +

{Q \over 2}

C +

{Q \over 4}

D

-

{Q \over 2}

Correct Answer

Option A

Solution

Force on + Q charge at x = 0 due to q charge, F1 =

{{KQq} \over {{{\left( {{d \over 2}} \right)}^2}}}

Force on +Q charge at x = 0 due to + Q charge at x = d is, F2 =

{{KQQ} \over {{d^2}}}

According to the question, F1 + F2 = 0 $\Rightarrow$ F1 = $-$ F2 $\Rightarrow$

{{KQq} \over {{{\left( {{d \over 2}} \right)}^2}}}

= $-$

{{KQQ} \over {{d^2}}}

$\Rightarrow$ 4q = $-$ Q $\Rightarrow$ q = $-$

{Q \over 4}