Electrostatics Questions — NEET Physics

Q1

An electric dipole with dipole moment

5 \times 10^{-6} \mathrm{Cm}

is aligned with the direction of a uniform electric field of magnitude

4 \times 10^5 \mathrm{~N} / \mathrm{C}

. The dipole is then rotated through an angle of

60^{\circ}

with respect to the electric field. The change in the potential energy of the dipole is:

A 1.2 J

B 1.5 J

C 0.8 J

D 1.0 J

Correct Answer

Option D

Solution

Given: Dipole moment, $|\vec{P}| = 5 \times 10^{-6} \, \text{Cm}$ Electric field magnitude, $|\vec{E}| = 4 \times 10^5 \, \text{N/C}$ Initial angle, $\theta_i = 0^\circ$ Final angle, $\theta_f = 60^\circ$ To calculate the change in potential energy $\Delta U$ of the dipole: $\Delta U = U_f - U_i = -PE \cos \theta_f + PE \cos \theta_i$ Simplifying, we have: $\Delta U = PE \left( \cos \theta_i - \cos \theta_f \right)$ Substitute the known values: $\Delta U = 5 \times 10^{-6} \times 4 \times 10^5 \left(1 - \dfrac{1}{2}\right)$ Calculate further: $\Delta U = 5 \times 10^{-6} \times 4 \times 10^5 \times \dfrac{1}{2}$ $\Delta U = 10 \times 10^{-6} \times 10^5$ Which simplifies to: $\Delta U = 1 \, \text{J}$ Thus, the change in the potential energy of the dipole is $\boxed{1 \, \text{J}}$ .

Q2

Two identical charged conducting spheres

A

and

B

have their centres separated by a certain distance. Charge on each sphere is

q

and the force of repulsion between them is

F

. A third identical uncharged conducting sphere is brought in contact with sphere

A

first and then with

B

and finally removed from both. New force of repulsion between spheres

A

and

B

(Radii of

A

and

B

are negligible compared to the distance of separation so that for calculating force between them they can be considered as point charges) is best given as:

A

\dfrac{F}{2}

B

\dfrac{3 F}{8}

C

\dfrac{3 F}{5}

D

\dfrac{2 F}{3}

Correct Answer

Option B

Solution

\begin{aligned} F^{\prime} & =\frac{\frac{K q}{2} \frac{3 q}{4}}{r^2} \\ F^{\prime} & =\frac{3 F}{8} \end{aligned}

Q3

A metal cube of side

5 \mathrm{~cm}

is charged with

6 \mu \mathrm{C}

. The surface charge density on the cube is

A

0.125 \times 10^{-3} \mathrm{C} \mathrm{m}^{-2}

B

0.25 \times 10^{-3} \mathrm{C} \mathrm{m}^{-2}

C

4 \times 10^{-3} \mathrm{C} \mathrm{m}^{-2}

D

0.4 \times 10^{-3} \mathrm{C} \mathrm{m}^{-2}

Correct Answer

Option D

Solution

To find the surface charge density of a charged metal cube, we need to determine the charge per unit area.

The surface charge density, denoted by $\sigma$ , is given by:

\sigma = \frac{Q}{A}

where: $Q$ is the total charge on the cube $A$ is the total surface area of the cube The cube has a side length of $5 \mathrm{~cm}$ , so each side of the cube is $5 \mathrm{~cm}$ .

Since 1 cm = 0.01 m, the side length in meters is:

5 \mathrm{~cm} = 0.05 \mathrm{~m}

The cube has 6 faces, and the area of one face is given by:

\text{Area of one face} = \left( 0.05 \mathrm{~m} \right)^2 = 0.0025 \mathrm{~m}^2

Therefore, the total surface area of the cube is:

A = 6 \times 0.0025 \mathrm{~m}^2 = 0.015 \mathrm{~m}^2

The total charge, $Q$ , is given as $6 \mu \mathrm{C}$ . Converting this to Coulombs:

6 \mu \mathrm{C} = 6 \times 10^{-6} \mathrm{~C}

Now, plugging the values into the formula for surface charge density:

\sigma = \frac{6 \times 10^{-6} \mathrm{~C}}{0.015 \mathrm{~m}^2} = 4 \times 10^{-4} \mathrm{~C} \mathrm{m}^{-2}

Therefore, the surface charge density on the cube is:

\sigma = 0.4 \times 10^{-3} \mathrm{~C} \mathrm{m}^{-2}

So, the correct answer is: Option D:

0.4 \times 10^{-3} \mathrm{C} \mathrm{m}^{-2}

Q4

The value of electric potential at a distance of

9 \mathrm{~cm}

from the point charge

4 \times 10^{-7} \mathrm{C}

is [Given

\dfrac{1}{4 \pi \varepsilon_0}=9 \times 10^9 \mathrm{~N} \mathrm{~m}^2 \mathrm{C}^{-2}

] :

A

4 \times 10^2 \mathrm{~V}

B

44.4 \mathrm{~V}

C

4.4 \times 10^5 \mathrm{~V}

D

4 \times 10^4 \mathrm{~V}

Correct Answer

Option D

Solution

The electric potential (V) at a distance (r) from a point charge (q) is given by:

V = \frac{1}{4 \pi \varepsilon_0} \frac{q}{r}

where

\frac{1}{4 \pi \varepsilon_0}

is Coulomb's constant. In this case, we have:

q = 4 \times 10^{-7} \mathrm{C}

r = 9 \mathrm{~cm} = 0.09 \mathrm{~m}

\frac{1}{4 \pi \varepsilon_0}=9 \times 10^9 \mathrm{~N} \mathrm{~m}^2 \mathrm{C}^{-2}

Substituting these values into the equation for electric potential, we get:

V = (9 \times 10^9 \mathrm{~N} \mathrm{~m}^2 \mathrm{C}^{-2}) \frac{(4 \times 10^{-7} \mathrm{C})}{(0.09 \mathrm{~m})}

V = 4 \times 10^4 \mathrm{~V}

Therefore, the value of electric potential at a distance of

9 \mathrm{~cm}

from the point charge

4 \times 10^{-7} \mathrm{C}

is

4 \times 10^4 \mathrm{~V}

. The correct answer is Option D.

Q5

A thin spherical shell is charged by some source. The potential difference between the two points

C

and

P

(in V) shown in the figure is: (Take

\dfrac{1}{4 \pi \varepsilon_0}=9 \times 10^9

SI units)

A

3 \times 10^5

B

1 \times 10^5

C

0.5 \times 10^5

D Zero

Correct Answer

Option D

Solution

For uniformly charged spherical shell,

\begin{aligned} & V=\frac{k q}{R} \quad(\text{ For } r \leq R) \\ & \therefore \quad V_C=V_P \\ & V_C-V_P=\text{ Zero } \\ \end{aligned}

Q6

According to Gauss law of electrostatics, electric flux through a closed surface depends on :

A the area of the surface

B the quantity of charges enclosed by the surface

C the shape of the surface

D the volume enclosed by the surface

Correct Answer

Option B

Solution

\phi=\frac{\mathrm{q}_{\text{inside }}}{\varepsilon_0}

only depends on charge enclosed by surface.

Q7

A charge

\mathrm{Q} ~\mu \mathrm{C}

is placed at the centre of a cube. The flux coming out from any one of its faces will be (in SI unit) :

A

\dfrac{Q}{\epsilon_0} \times 10^{-6}

B

\dfrac{2 \mathrm{Q}}{3 \epsilon_0} \times 10^{-3}

C

\dfrac{\mathrm{Q}}{6 \epsilon_0} \times 10^{-3}

D

\dfrac{\mathrm{Q}}{6 \epsilon_0} \times 10^{-6}

Correct Answer

Option D

Solution

The Gaussian surface for the charge placed at the center of the cube will spread out equally through all sides of the cube.

According to Gauss's Law, electric flux

\Phi

through a closed surface is equal to the charge enclosed

Q

divided by the permittivity of free space

\epsilon_0

:

\Phi_{\text{total}} = \frac{Q}{\epsilon_0}

Given that the cube has 6 faces, and due to symmetry, the flux through each face will be equal, the flux through any one face

\Phi_{\text{face}}

is:

\Phi_{\text{face}} = \frac{\Phi_{\text{total}}}{6} = \frac{Q}{6\epsilon_0}

Since the charge is given in microcoulombs (

\mu\mathrm{C}

), we need to convert it to coulombs by multiplying with

10^{-6}

:

\Phi_{\text{face}} = \frac{Q \times 10^{-6}}{6\epsilon_0}

This matches option D, which means the correct flux through one face of the cube given a charge Q microcoulombs at the center is

\frac{Q}{6\epsilon_0} \times 10^{-6}

.

Q8

If a conducting sphere of radius

\mathrm{R}

is charged. Then the electric field at a distance

\mathrm{r}(\mathrm{r} > \mathrm{R})

from the centre of the sphere would be,

(\mathrm{V}=

potential on the surface of the sphere)

A

\dfrac{r V}{R^2}

B

\dfrac{R^2 V}{r^3}

C

\dfrac{R V}{r^2}

D

\dfrac{\mathrm{V}}{\mathrm{r}}

Correct Answer

Option C

Solution

\begin{aligned} & \therefore V=\frac{K Q}{R} \\ & E=\frac{K Q}{r^2} \\ & E=\frac{V R}{r^2} \end{aligned}

Q9

An electric dipole is placed at an angle of

30^{\circ}

with an electric field of intensity

2 \times 10^{5} \mathrm{NC}^{-1}

. It experiences a torque equal to

4 ~\mathrm{N~m}

. Calculate the magnitude of charge on the dipole, if the dipole length is

2 \mathrm{~cm}

.

A 6 mC

B 4 mC

C 2 mC

D 8 mC

Correct Answer

Option C

Solution

The torque τ experienced by an electric dipole in an electric field is given by the formula:

\tau = pE\sin{\theta}

where p is the electric dipole moment, E is the electric field intensity, and θ is the angle between the dipole and the electric field.

The electric dipole moment p can be expressed as:

p = qd

where q is the charge on the dipole, and d is the dipole length.

We are given the following values: Torque τ = 4 N·m Electric field intensity E =

2 \times 10^5 ~\mathrm{NC}^{-1}

Angle θ =

30^{\circ}

Dipole length d = 2 cm = 0.02 m We need to find the charge q on the dipole.

Let's first solve for the electric dipole moment p:

\tau = pE\sin{\theta}

$\Rightarrow$

p = \frac{\tau}{E\sin{\theta}}

Substituting the given values:

p = \frac{4}{(2 \times 10^5) \sin{30^{\circ}}} = \frac{4}{(2 \times 10^5)(0.5)} = \frac{4}{10^5} = 4 \times 10^{-5} ~\mathrm{C~m}

Now, let's solve for the charge q using the formula: $\Rightarrow$

p = qd

q = \frac{p}{d}

Substituting the values for p and d:

q = \frac{4 \times 10^{-5}}{0.02} = 2 \times 10^{-3} ~\mathrm{C} = 2 ~\mathrm{mC}

So, the magnitude of the charge on the dipole is 2 mC.

Q10

If

\oint\limits_{s} \vec{E} \cdot \overrightarrow{d S}=0

over a surface, then:

A the magnitude of electric field on the surface is constant.

B all the charges must necessarily be inside the surface.

C the electric field inside the surface is necessarily uniform.

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

Correct Answer

Option D

Solution

\phi_{\mathrm{closed}}=0

So,

\mathrm{\phi_{in}=\phi_{out}}

Number of field lines entering is equal number of field lines leaving.