Electrostatics — Page 7 | NEET Physics

Q61

Two charges q 1 and q 2 are placed 30 cm apart, as shown in the figure. A third charge q 3 is moved along the arc of a circle of radius 40 cm from C to D. The change in the potential energy of the system is

{{{q_3}} \over {4\pi {\varepsilon _0}}}

where k is

A 8q 1

B 6q 1

C 8q 2

D 6q 2

Correct Answer

Option C

Solution

We know that potential energy of discrete system of charges is given by

U = {1 \over {4\pi {\varepsilon _0}}}\left( {{{{q_1}{q_2}} \over {{r_{12}}}} + {{{q_2}{q_3}} \over {{r_{23}}}} + {{{q_3}{q_1}} \over {{r_{31}}}}} \right)

According to question,

{U_{initial}} = {1 \over {4\pi {\varepsilon _0}}}\left( {{{{q_1}{q_2}} \over {0.3}} + {{{q_2}{q_3}} \over {0.5}} + {{{q_3}{q_1}} \over {0.4}}} \right)

{U_{final}} = {1 \over {4\pi {\varepsilon _0}}}\left( {{{{q_1}{q_2}} \over {0.3}} + {{{q_2}{q_3}} \over {0.1}} + {{{q_3}{q_1}} \over {0.4}}} \right)

{U_{final}} - {U_{initial}} = {1 \over {4\pi {\varepsilon _0}}}\left( {{{{q_1}{q_2}} \over {0.1}} - {{{q_2}{q_3}} \over {0.5}}} \right)

= {1 \over {4\pi {\varepsilon _0}}}\left[ {10{q_2}{q_3} - 2{q_2}{q_3}} \right] = {{{q_3}} \over {4\pi {\varepsilon _0}}}\left( {8{q_2}} \right)

Q62

A bullet of mass 2 g is having a charge of 2

\mu

C. Through what potential difference must it be accelerated, starting from rst, to acquire a speed of 10 m/s ?

A 5 kV

B 50 kV

C 5 V

D 50 V

Correct Answer

Option B

Solution

Using

{1 \over 2}m{v^2} = qV

V = {1 \over 2} \times {{2 \times {{10}^{ - 3}} \times 10 \times 10} \over {2 \times {{10}^{ - 6}}}} = 50\,kV

Q63

An electric dipole has the magnitude of its charge as q and its dipole moment is p. It is placed in a uniform electric field E. If its dipole moment is along the direction of the field, the force on it and its potential energy are respectively

A 2q.E and minimum

B q.E and p.E

C zero and minimum

D q.E and maximum

Correct Answer

Option C

Solution

When the dipole is in the direction of field then net force is qE + (–qE) = 0 and its potential energy is minimum = – P.E. = –qaE

Q64

A charge q is located at the centre of a cube. The electric flux through any face is

A

{{2\pi q} \over {6\left( {4\pi {\varepsilon _0}} \right)}}

B

{{4\pi q} \over {6\left( {4\pi {\varepsilon _0}} \right)}}

C

{{\pi q} \over {6\left( {4\pi {\varepsilon _0}} \right)}}

D

{q \over {6\left( {4\pi {\varepsilon _0}} \right)}}

Correct Answer

Option B

Solution

The total flux through the cube

{\phi _{total}} = {q \over {{\varepsilon _0}}}

$\therefore$ The electric flux through any face

{\phi _{face}} = {q \over {6{\varepsilon _0}}} = {{4\pi q} \over {6\left( {4\pi {\varepsilon _0}} \right)}}

Q65

Identical charges (

-

q) are placed at each corners of cube of side b then electrostatic potential energy of charge (+q) which is placed at centre of cube will be

A

{{ - 4\sqrt 2 {q^2}} \over {\pi {\varepsilon _0}b}}

B

{{ - 8\sqrt 2 {q^2}} \over {\pi {\varepsilon _0}b}}

C

{{ - 4\,{q^2}} \over {\sqrt 3 \,\pi {\varepsilon _0}b}}

D

{{8\sqrt 2 \,{q^2}} \over {4\,\pi {\varepsilon _0}b}}

Correct Answer

Option C

Solution

There are eight corners of a cube and in each corner there is a charge of (–q).

At the centre of the corner there is a charge of (+q).

Each corner is equidistant from the centres of the cube and the distance (d) is half of the diagonals of the cube.

Diagonal of the cube =

\sqrt {{b^2} + {b^2} + {b^2}} = \sqrt 3 b

$\therefore$

d = \sqrt 3 b/2

Now, electric potential energy of the charge (+q) due to a charge (–q) at one corner U

= {{{q_1}{q_2}} \over {4\pi {\varepsilon _0}r}} = {{\left( { + q} \right) \times \left( { - q} \right)} \over {4\pi {\varepsilon _0}\left( {\sqrt 3 b/2} \right)}} = - {{{q^2}} \over {2\pi {\varepsilon _0}\left( {\sqrt 3 b} \right)}}

$\therefore$ Total electric potential energy due to all the eight identical charges =

8U = - {{8{q^2}} \over {2\pi {\varepsilon _0}\sqrt 3 b}} = {{ - 4{q^2}} \over {\sqrt 3 \pi {\varepsilon _0}b}}

Q66

Some charge is being given to a conductor. Then its potential is

A maximum at surface

B maximum at centre

C remain same throughout the conductor

D maximum somewhere between surface and centre.

Correct Answer

Option C

Solution

Electric field intensity E is zero within a conductor due to charge given to it. Also,

E = - {{dV} \over {dx}} \Rightarrow {{dV} \over {dx}} = 0

(inside the conductor) $\therefore$ V = constant. [V is potential] So potential remains same throughout the conductor.

Q67

A charge Q

\mu

C is placed at the centre of a cube, the flux coming out from each face will be

A

{Q \over {6{\varepsilon _0}}} \times {10^{ - 6}}

B

{Q \over {6{\varepsilon _0}}} \times {10^{ - 3}}

C

{Q \over {24{\varepsilon _0}}}

D

{Q \over {8{\varepsilon _0}}}

Correct Answer

Option A

Solution

For complete cube

\phi = {Q \over {{\varepsilon _0}}} \times {10^{ - 6}}

For each face

\phi = {1 \over 6}{Q \over {{\varepsilon _0}}} \times {10^{ - 6}}

Q68

A dipole of dipole moment

\overrightarrow p

is placed in uniform electric field

\overrightarrow E

then torque acting on it is given by

A

\overrightarrow \tau = \overrightarrow p .\overrightarrow E

B

\overrightarrow \tau = \overrightarrow p \times \overrightarrow E

C

\overrightarrow \tau = \overrightarrow p + \overrightarrow E

D

\overrightarrow \tau = \overrightarrow p - \overrightarrow E

Correct Answer

Option B

Solution

When an electric dipole is placed in a uniform electrical field

\overrightarrow E

, the torque on the dipole is given by

\overrightarrow \tau = \overrightarrow p \times \overrightarrow E

Q69

A charge Q is situated at the corner of a cube, the electric flux passed through all the six faces of the cube is

A

{Q \over {6{\varepsilon _0}}}

B

{Q \over {8{\varepsilon _0}}}

C

{Q \over {{\varepsilon _0}}}

D

{Q \over {2{\varepsilon _0}}}

Correct Answer

Option B

Solution

Using Gauss’s law, the total electric flux through a closed surface is

\Phi = \frac{q_{\text{enclosed}}}{\varepsilon_0}

Here, the charge $Q$ is placed at a corner of the cube, not at the centre.

To apply Gauss’s law easily, imagine $8$ identical cubes joined together so that the charge $Q$ comes at the common corner and becomes the centre of the larger cube.

Now for the larger cube: the charge enclosed = $Q$ so total flux through the larger cube is

\Phi_{\text{large cube}} = \frac{Q}{\varepsilon_0}

Since the larger cube is made of $8$ identical small cubes, by symmetry the flux is equally shared by them.

So, flux through one small cube is

\Phi_{\text{one cube}} = \frac{1}{8}\cdot \frac{Q}{\varepsilon_0} = \frac{Q}{8\varepsilon_0}

Hence, the electric flux through all the six faces of the given cube is

\boxed{\frac{Q}{8\varepsilon_0}}

So, the correct answer is:

\boxed{\text{Option B}}

Q70

Electric field at centre O of semicircle of radius

a

having linear charge density

\lambda

given as

A

{{2\lambda } \over {{\varepsilon _0}a}}

B

{{\lambda \pi } \over {{\varepsilon _0}a}}

C

{\lambda \over {2\pi {\varepsilon _0}a}}

D

{\lambda \over {\pi {\varepsilon _0}a}}

Correct Answer

Option C

Solution

Considering symmetric elements each of length dl at A and B, we note that electric fields perpendicular to PO are cancelled and those along PO are added.

The electric field due to an element of length

dl( = ad\theta)

along PO.

dE = {1 \over {4\pi {\varepsilon _0}}}{{dq} \over {{a^2}}}\cos \theta

\left(\because {dl = ad\theta } \right)

= {1 \over {4\pi {\varepsilon _0}}}{{\lambda dl} \over {{a^2}}}\cos \theta

= {1 \over {4\pi {\varepsilon _0}}}{{\lambda \left( {ad\theta } \right)} \over {{a^2}}}\cos \theta

Net electric field at O

E = \int_{ - \pi /2}^{\pi /2} {dE}

= 2\int_O^{\pi /2} {{1 \over {4\pi {\varepsilon _0}}}{{\lambda \left( {ad\theta } \right)} \over {{a^2}}}\cos \theta }

= 2.{1 \over {4\pi {\varepsilon _0}}}{\lambda \over a}\left[ {\sin \theta } \right]_o^{\pi /2}

= 2.{1 \over {4\pi {\varepsilon _0}}}{\lambda \over a}.1 = {\lambda \over {2\pi {\varepsilon _0}a}}