Electrostatics — Page 8 | JEE Physics

Q71

Two identical metallic spheres

\mathrm{A}

and

\mathrm{B}

when placed at certain distance in air repel each other with a force of

\mathrm{F}

. Another identical uncharged sphere

\mathrm{C}

is first placed in contact with

\mathrm{A}

and then in contact with

\mathrm{B}

and finally placed at midpoint between spheres A and B. The force experienced by sphere C will be:

A 3F/2

B 3F/4

C F

D 2F

Correct Answer

Option B

Solution

Let two identical spheres have charge q. And distance between them = r $\therefore$ Force between the spheres

(F) = {{k{q^2}} \over {{r^2}}}

Now when an identical uncharged sphere C comes in contact with A, charge q on sphere A get's divided equally to both sphere.

So, both sphere A and C have charge

= {q \over 2}

Now, C get's in contact with B. So their total charge

\left( {q + {q \over 2}} \right)

gets divided equally. So, charge on both B and C is

= {{q + {q \over 2}} \over 2} = {{3q} \over 4}

Now, C is place midpoint between A and B. Repulsion force between A and C,

{F_{AC}} = {{k\left( {{q \over 2}} \right)\left( {{{3q} \over 4}} \right)} \over {{{\left( {{r \over 2}} \right)}^2}}} = {{4k \times 3{q^2}} \over {8{r^2}}}

Repulsion force between B and C,

{F_{BC}} = {{k\left( {{{3q} \over 4}} \right)\left( {{{3q} \over 4}} \right)} \over {{{\left( {{r \over 2}} \right)}^2}}} = {{4k \times 9{q^2}} \over {16{r^2}}}

$\therefore$ Net force on C,

{F_{net}} = {F_{BC}} - {F_{AC}}

= {{4k \times 9{q^2}} \over {16{r^2}}} - {{4k \times 3{q^2}} \over {8{r^2}}}

= {{k{q^2}} \over {{r^2}}}\left( {{9 \over 4} - {3 \over 2}} \right)

= {{k{q^2}} \over {{r^2}}} \times {3 \over 4}

= {3 \over 4}F

[as

F = {{k{q^2}} \over {{r^2}}}

]

Q72

Two isolated metallic solid spheres of radii

\mathrm{R}

and

2 \mathrm{R}

are charged such that both have same charge density

\sigma

. The spheres are then connected by a thin conducting wire. If the new charge density of the bigger sphere is

\sigma^{\prime}

. The ratio

\dfrac{\sigma^{\prime}}{\sigma}

is :

A

\dfrac{5}{3}

B

\dfrac{5}{6}

C

\dfrac{9}{4}

D

\dfrac{4}{3}

Correct Answer

Option B

Solution

\sigma = {{{Q_1}} \over {4\pi {R^2}}} = {{{Q_2}} \over {4\pi {{(2R)}^2}}}

Now

Q{'_2} = {Q_{total}}\left[ {{{{R_2}} \over {{R_1} + {R_2}}}} \right]

= ({Q_1} + {Q_2})\left[ {{{2R} \over {3R}}} \right]

= \sigma (20\pi {R^2}){2 \over 3}

$\therefore$

\sigma {'_2} = {{Q{'_2}} \over {4\pi {{(2R)}^2}}} = {{\sigma (20\pi {R^2}){2 \over 3}} \over {16\pi {R^2}}}

= {5 \over 4} \times {2 \over 3}\sigma

= {5 \over 6}\sigma

Q73

A point charge

2\times10^{-2}~\mathrm{C}

is moved from P to S in a uniform electric field of

30~\mathrm{NC^{-1}}

directed along positive x-axis. If coordinates of P and S are (1, 2, 0) m and (0, 0, 0) m respectively, the work done by electric field will be

A 600 mJ

B

-1200

mJ

C 1200 mJ

D

-600

mJ

Correct Answer

Option D

Solution

w = \int {F\,.\,ds}

\overrightarrow F = q\overrightarrow E = 2 \times {10^{ - 2}} \times 30\widehat i = 0.6N\widehat i

w = \overrightarrow F .\overrightarrow d = (0.6\widehat i)\,.\,( - \widehat i, - 2\widehat j)

= - 0.6

J

= - 600

mJ

Q74

In a cuboid of dimension

2 \mathrm{~L} \times 2 \mathrm{~L} \times \mathrm{L}

, a charge

q

is placed at the center of the surface '

\mathrm{S}

' having area of

4 \mathrm{~L}^{2}

. The flux through the opposite surface to '

\mathrm{S}

' is given by

A

\dfrac{q}{2 \epsilon_{0}}

B

\dfrac{q}{3 \epsilon_{0}}

C

\dfrac{q}{12 \epsilon_{0}}

D

\dfrac{q}{6 \in_{0}}

Correct Answer

Option D

Solution

If we consider a similar box above this box then it becomes cube of side length $2 L$ $\phi$ through a surface $=\dfrac{q}{6 \varepsilon_{0}}$

Q75

If two charges q

_1

and q

_2

are separated with distance 'd' and placed in a medium of dielectric constant K. What will be the equivalent distance between charges in air for the same electrostatic force?

A

d\sqrt k

B

1\,.\,5d\sqrt k

C

k\sqrt d

D

2d\sqrt k

Correct Answer

Option A

Solution

\begin{aligned} & \text{ dielectric constant }=\mathrm{K} \\\\ & F_{\text{medium }}=\frac{1}{4 \pi\left(\mathrm{K} \varepsilon_{0}\right)} \times \frac{q_{1} q_{2}}{d^{2}} \\\\ & \because \quad F_{\text{air }}=F_{\text{medium }} \\\\ & \frac{1}{4 \pi \varepsilon_{0}} \frac{q_{1} q_{2}}{\left(d_{\mathrm{air}}\right)^{2}}=\frac{1}{4 \pi\left(\mathrm{K} \varepsilon_{0}\right)} \frac{q_{1} q_{2}}{d^{2}} \\\\ & \therefore \quad d_{\text{air }}=d\sqrt{\mathrm{K}} \end{aligned}

Q76

The electric field due to a short electric dipole at a large distance

(r)

from center of dipole on the equatorial plane varies with distance as :

A

\dfrac{1}{r^{2}}

B

\dfrac{1}{r}

C

r

D

\dfrac{1}{r^{3}}

Correct Answer

Option D

Solution

At a large distance $r$ from the center of a short electric dipole, the electric field on the equatorial plane can be approximated as:

E = \frac{1}{4\pi\epsilon_0}\frac{2p}{r^3}

where $p$ is the dipole moment of the electric dipole, and $\epsilon_0$ is the permittivity of free space.

This formula is derived using the concept of electric dipole moment, which is defined as:

\vec{p} = q\vec{d}

where $q$ is the magnitude of the electric charge, and $\vec{d}$ is the separation vector between the positive and negative charges of the dipole.

The electric field at a point on the equatorial plane of the dipole is due to the electric field of the positive and negative charges at that point.

Since the charges are equal in magnitude and opposite in sign, their electric fields at a point on the equatorial plane cancel out along the axis of the dipole, leaving only the component perpendicular to the axis.

This perpendicular component of the electric field is proportional to the dipole moment $p$ and inversely proportional to the cube of the distance $r$ from the center of the dipole.

Therefore, the electric field due to a short electric dipole at a large distance $r$ from the center of the dipole on the equatorial plane varies with distance as:

\boxed{E \propto \frac{1}{r^3}}

where the proportionality constant is $\dfrac{1}{4\pi\epsilon_0}$ .

Q77

Two charges each of magnitude

0.01 ~\mathrm{C}

and separated by a distance of

0.4 \mathrm{~mm}

constitute an electric dipole. If the dipole is placed in an uniform electric field '

\vec{E}

' of 10 dyne/C making

30^{\circ}

angle with

\vec{E}

, the magnitude of torque acting on dipole is:

A

4 \cdot 0 \times 10^{-10} ~\mathrm{Nm}

B

1.5 \times 10^{-9} ~\mathrm{Nm}

C

1.0 \times 10^{-8} ~\mathrm{Nm}

D

2.0 \times 10^{-10} ~\mathrm{Nm}

Correct Answer

Option D

Solution

Given two charges each of magnitude

0.01 \,\text{C}

and separated by a distance of

0.4 \,\text{mm} = 0.4 \times 10^{-3} \,\text{m}

, we can calculate the dipole moment

\vec{p}

:

\vec{p} = q \cdot \vec{d} = (0.01 \,\text{C})(0.4 \times 10^{-3} \,\text{m}) = 4 \times 10^{-6} \,\text{Cm}

The dipole is placed in a uniform electric field

\vec{E}

of magnitude

10 \,\text{dyne/C}

, which is equivalent to

10 \times 10^{-5} \,\text{N/C}

. The angle between the dipole moment and the electric field is

30^{\circ}

. The torque acting on the dipole can be calculated using the formula:

\tau = pE \sin{\theta}

Substituting the given values:

\tau = (4 \times 10^{-6} \,\text{Cm})(10 \times 10^{-5} \,\text{N/C})\sin{30^{\circ}}

\tau = (4 \times 10^{-6} \,\text{Cm})(10^{-4} \,\text{N/C})(\frac{1}{2})

\tau = 2 \times 10^{-10} \,\text{Nm}

The magnitude of the torque acting on the dipole is

2.0 \times 10^{-10} \,\text{Nm}

.

Q78

Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R Assertion A : If an electric dipole of dipole moment

30 \times 10^{-5} ~\mathrm{C} ~\mathrm{m}

is enclosed by a closed surface, the net flux coming out of the surface will be zero. Reason R : Electric dipole consists of two equal and opposite charges. In the light of above, statements, choose the correct answer from the options given below.

A A is true but R is false

B A is false but R is true

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

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

Correct Answer

Option D

Solution

Assertion A: If an electric dipole of dipole moment

30 \times 10^{-5} ~\mathrm{C} ~\mathrm{m}

is enclosed by a closed surface, the net flux coming out of the surface will be zero.

This statement is true.

According to Gauss's Law, the electric flux through a closed surface is proportional to the net charge enclosed by the surface.

Since an electric dipole consists of two equal and opposite charges, the net charge enclosed by the surface is zero, and therefore, the net flux coming out of the surface will also be zero.

Reason R: Electric dipole consists of two equal and opposite charges.

This statement is also true.

An electric dipole is formed by two equal and opposite charges separated by a fixed distance.

The reason R correctly explains the assertion A because the net charge enclosed by the surface is zero due to the presence of equal and opposite charges in the electric dipole, resulting in zero net electric flux coming out of the surface.

Q79

Electric potential at a point '

\mathrm{P}

' due to a point charge of

5 \times 10^{-9} \mathrm{C}

is

50 \mathrm{~V}

. The distance of '

\mathrm{P}

' from the point charge is: (Assume,

\dfrac{1}{4 \pi \varepsilon_{0}}=9 \times 10^{+9} ~\mathrm{Nm}^{2} \mathrm{C}^{-2}

)

A 0.9 cm

B 90 cm

C 3 cm

D 9 cm

Correct Answer

Option B

Solution

The electric potential (V) at a distance (r) from a point charge (Q) is given by the formula: $V = \dfrac{1}{4\pi\epsilon_0} \dfrac{Q}{r}$ In this case, we know (V) and (Q), and we're asked to solve for (r).

We can rearrange the formula to solve for (r): $r = \dfrac{1}{4\pi\epsilon_0} \dfrac{Q}{V}$ Substituting the given values into this equation gives: $r = \dfrac{9 \times 10^9 \, \text{Nm}^2 \text{C}^{-2}}{1} \dfrac{5 \times 10^{-9} \, \text{C}}{50 \, \text{V}} = 0.9 \, \text{m}$ So, the distance of the point P from the point charge is 0.9 meters.

This corresponds to 90 cm.

Q80

A dipole comprises of two charged particles of identical magnitude

q

and opposite in nature. The mass 'm' of the positive charged particle is half of the mass of the negative charged particle. The two charges are separated by a distance '

l

'. If the dipole is placed in a uniform electric field '

\bar{E}

'; in such a way that dipole axis makes a very small angle with the electric field, '

\bar{E}

'. The angular frequency of the oscillations of the dipole when released is given by:

A

\sqrt{\dfrac{3 q E}{2 m l}}

B

\sqrt{\dfrac{4 q E}{m l}}

C

\sqrt{\dfrac{8 q E}{3 m l}}

D

\sqrt{\dfrac{8 q E}{m l}}

Correct Answer

Option A

Solution

If released, it will oscillate about centre of mass. For small ' $\theta$ '

\begin{aligned} & \tau=-\mathrm{PE} . \theta \\\\ & \Rightarrow\left[2 \mathrm{~m} \frac{1^2}{9}+\mathrm{m} \frac{4 \mathrm{l}^2}{9}\right] \alpha=-\mathrm{qlE} \cdot \theta \\\\ & \Rightarrow \frac{2 \mathrm{ml}^2}{3} \alpha=-\mathrm{qlE} \cdot \theta \Rightarrow \alpha=-\frac{3 \mathrm{qE}}{2 \mathrm{ml}} \theta \\\\ & \omega=\sqrt{\frac{3 \mathrm{qE}}{2 \mathrm{ml}}} \end{aligned}