Electrostatics — Page 9 | JEE Physics

Q81

An electric charge

10^{-6} \mu \mathrm{C}

is placed at origin

(0,0)

\mathrm{m}

of

\mathrm{X}-\mathrm{Y}

co-ordinate system. Two points

\mathrm{P}

and

\mathrm{Q}

are situated at

(\sqrt{3}, \sqrt{3}) \mathrm{m}

and

(\sqrt{6}, 0) \mathrm{m}

respectively. The potential difference between the points

\mathrm{P}

and

\mathrm{Q}

will be :

A

\sqrt{3} \mathrm{~V}

B

\sqrt{6} \mathrm{~V}

C

0 \mathrm{~V}

D

3 \mathrm{~V}

Correct Answer

Option C

Solution

Potential difference

=\frac{K Q}{r_1}-\frac{K Q}{r_2}

\begin{aligned} & r_1=\sqrt{(\sqrt{3})^2+(\sqrt{3})^2} \\ & r_2=\sqrt{(\sqrt{6})^2+0} \end{aligned}

As

r_1=r_2=\sqrt{6} \mathrm{~m}

So potential difference

=0

Q82

Given below are two statements : one is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A) : Work done by electric field on moving a positive charge on an equipotential surface is always zero. Reason (R) : Electric lines of forces are always perpendicular to equipotential surfaces. 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)

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

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

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

Correct Answer

Option A

Solution

The most appropriate answer from the options given would be Option A: Both (A) and (R) are correct, and (R) is the correct explanation of (A).

Here is the reasoning for this answer: Assertion (A) states that the work done by an electric field on moving a positive charge on an equipotential surface is always zero.

This statement is true because by definition, an equipotential surface is a surface over which the electric potential is constant.

When a charge moves along an equipotential surface, there is no change in its electric potential energy since potential difference

\Delta V

is zero. Work done (

W

) is defined as the product of charge (

q

), potential difference (

\Delta V

), and the cosine of the angle between the field and direction of motion (

\cos \theta

), which can be written as:

W = q \Delta V \cos \theta

Because

\Delta V = 0

on an equipotential surface, irrespective of the value of

\cos \theta

, the work

W

will be zero.

Hence, the Assertion (A) is correct.

Reason (R) says that electric lines of forces are always perpendicular to equipotential surfaces.

This statement is also correct as the electric field lines, by definition, are directed such that they are tangent to the electric field vector at any point in space.

Since the electric potential is constant on an equipotential surface, there can be no component of the electric field parallel to the surface, as that would imply a force and potential change along the surface.

The electric field thus must be perpendicular to the equipotential surface, which means that the electric field lines must also be perpendicular to the equipotential surface.

In other words, the electric field does no work when a charge moves along an equipotential surface because the motion is perpendicular to the force.

Therefore, Reason (R) is not only correct, but it is also the correct explanation for Assertion (A), making Option A the right choice.

Q83

Force between two point charges

q_1

and

q_2

placed in vacuum at '

r

' cm apart is

F

. Force between them when placed in a medium having dielectric constant

K=5

at '

r / 5

'

\mathrm{cm}

apart will be:

A

5 F

B

25 F

C

F / 5

D

F / 25

Correct Answer

Option A

Solution

In air

F=\frac{1}{4 \pi \epsilon_0} \frac{\mathrm{q}_1 \mathrm{q}_2}{\mathrm{r}_2}

In medium

\mathrm{F}^{\prime}=\frac{1}{4 \pi\left(\mathrm{K} \epsilon_0\right)} \frac{\mathrm{q}_1 \mathrm{q}_2}{\left(\mathrm{r}^{\prime}\right)^2}=\frac{25}{4 \pi\left(5 \epsilon_0\right)} \frac{\mathrm{q}_1 \mathrm{q}_2}{(\mathrm{r})^2}=5 \mathrm{~F}

Q84

An electric field is given by

(6 \hat{i}+5 \hat{j}+3 \hat{k}) \mathrm{N} / \mathrm{C}

. The electric flux through a surface area

30 \hat{i} \mathrm{~m}^2

lying in YZ-plane (in SI unit) is :

A 60

B 90

C 180

D 150

Correct Answer

Option C

Solution

\begin{aligned} & \overrightarrow{\mathrm{E}}=6 \hat{\mathrm{i}}+5 \hat{\mathrm{j}}+3 \hat{\mathrm{k}} \\ & \overrightarrow{\mathrm{A}}=30 \hat{\mathrm{i}} \\ & \phi=\overrightarrow{\mathrm{E}} \cdot \overrightarrow{\mathrm{A}} \\ & \phi=(6 \hat{\mathrm{i}}+5 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}) \cdot(30 \hat{\mathrm{i}}) \\ & \phi=6 \times 30=180 \end{aligned}

Q85

A particle of charge '

-q

' and mass '

m

' moves in a circle of radius '

r

' around an infinitely long line charge of linear charge density '

+\lambda

'. Then time period will be given as : (Consider

k

as Coulomb's constant)

A

T^2=\dfrac{4 \pi^2 m}{2 k \lambda q} r^3

B

T=\dfrac{1}{2 \pi r} \sqrt{\dfrac{m}{2 k \lambda q}}

C

T=\dfrac{1}{2 \pi} \sqrt{\dfrac{2 k \lambda q}{m}}

D

T=2 \pi r \sqrt{\dfrac{m}{2 k \lambda q}}

Correct Answer

Option D

Solution

\begin{aligned} & \frac{2 \mathrm{k} \lambda \mathrm{q}}{\mathrm{r}}=\mathrm{m} \omega^2 \mathrm{r} \\ & \omega^2=\frac{2 \mathrm{k} \lambda \mathrm{q}}{\mathrm{mr}^2} \\ & \left(\frac{2 \pi}{\mathrm{T}}\right)^2=\frac{2 \mathrm{k} \lambda \mathrm{q}}{\mathrm{mr}^2} \\ & \mathrm{~T}=2 \pi \mathrm{r} \sqrt{\frac{\mathrm{m}}{2 \mathrm{k} \lambda \mathrm{q}}} \end{aligned}

Q86

The electrostatic potential due to an electric dipole at a distance '

r

' varies as :

A

\dfrac{1}{r^3}

B

\dfrac{1}{\mathrm{r}}

C

\dfrac{1}{r^2}

D r

Correct Answer

Option C

Solution

V=\frac{k P \cos \theta}{r^2}

& can also checked dimensionally

Q87

A charge

q

is placed at the center of one of the surface of a cube. The flux linked with the cube is:

A

\dfrac{q}{2 \epsilon_0}

B Zero

C

\dfrac{q}{4 \epsilon_0}

D

\dfrac{q}{8 \epsilon_0}

Correct Answer

Option A

Solution

When considering the electric flux linked with a cube due to a charge placed at one of its surfaces, it's important to apply Gauss's law.

Gauss's law states that the total electric flux through a closed surface is equal to

\frac{q_{\text{enc}}}{\epsilon_0}

, where

q_{\text{enc}}

is the charge enclosed by the surface and

\epsilon_0

is the permittivity of free space. In the given scenario, the charge

q

is placed at the center of one of the surfaces of the cube.

Conceptually, we can think of this arrangement as part of a larger situation where if we had a larger, imaginary cube that encompasses the entire setup such that the point charge is at its geometric center, the charge would then be uniformly distributing its flux through all six faces of this larger cube.

However, since the actual setup involves only one cube with one face adjacent to the charge, we essentially have only one-half of the total possible geometry through which the flux from this charge can emerge - implying the flux through our actual cube is a fraction of the total flux that would emanate from the charge if it were centrally placed within a larger, encompassing cube.

Therefore, only half of the flux emanating from the charge will pass through the actual cube because the charge is placed directly on one of its surfaces, effectively distributing its influence through 180 degrees (half of the space around it) instead of the full 360 degrees.

Hence, the flux linked with the cube will be half of the total flux

\frac{q}{\epsilon_0}

, which is calculated using Gauss's law for a point charge. Therefore, the correct answer is

\frac{q}{2\epsilon_0}

. Option A

\frac{q}{2 \epsilon_0}

is the correct choice.

Q88

Two charged conducting spheres of radii

a

and

b

are connected to each other by a conducting wire. The ratio of charges of the two spheres respectively is:

A

a b

B

\dfrac{b}{a}

C

\dfrac{a}{b}

D

\sqrt{a b}

Correct Answer

Option C

Solution

When two conducting spheres of radii

a

and

b

are connected by a conducting wire, they come to the same potential because conductors in contact share charges until their potentials become equal.

The potential of a charged sphere is given by

V = \frac{kQ}{R}

, where

V

is the potential,

k

is Coulomb's constant,

Q

is the charge on the sphere, and

R

is the radius of the sphere. For the two spheres at the same potential, we have:

\frac{kQ_a}{a} = \frac{kQ_b}{b}

Where:

Q_a

and

Q_b

are the charges on the spheres with radii

a

and

b

, respectively.

k

cancels out from both sides as it is a constant. From the above equation, to find the ratio of charges

\frac{Q_a}{Q_b}

, we rearrange it as follows:

\frac{Q_a}{Q_b} = \frac{a}{b}

Therefore, the ratio of the charges of the two spheres respectively is

\frac{a}{b}

. So, the correct answer is Option C:

\frac{a}{b}

.

Q89

The vehicles carrying inflammable fluids usually have metallic chains touching the ground:

A To protect tyres from catching dirt from ground

B It is a custom

C To alert other vehicles

D To conduct excess charge due to air friction to ground and prevent sparking

Correct Answer

Option D

Solution

The correct option is Option D: To conduct excess charge due to air friction to ground and prevent sparking.

This method is grounded in the principles of physics, particularly relating to static electricity and grounding.

As vehicles move through the air, especially at high speeds, friction between the air and the vehicle can lead to the accumulation of static electricity on the vehicle.

This is a common phenomenon and is more pronounced in dry conditions where humidity is low, as moisture in the air can help dissipate static charge more effectively.

In the case of vehicles carrying inflammable fluids, the presence of static electricity poses a significant risk.

This is because a static discharge (or sparking) in the presence of flammable vapors can ignite those vapors, leading to potentially catastrophic fires or explosions.

The metallic chains that touch the ground serve a crucial safety function by providing a path for the static electrical charge to safely dissipate into the earth, a process known as grounding.

The earth acts as a vast reservoir that can absorb large amounts of electrical charge.

By grounding the vehicle in this way, the risk of static discharge into the surrounding atmosphere is significantly reduced, enhancing safety by preventing ignition of inflammable materials.

The effectiveness of this safety measure hinges on the electrical conductivity of the chain and its contact with the ground.

The chain must be made of a material with good electrical conductivity (such as metal) and maintain adequate contact with the ground to ensure that the static charge can be continuously dissipated.

This safety precaution is a practical application of electrostatic principles, showcasing how understanding and harnessing the laws of physics can provide effective solutions to real-world problems.

Q90

Two identical conducting spheres P and S with charge Q on each, repel each other with a force

16 \mathrm{~N}

. A third identical uncharged conducting sphere

\mathrm{R}

is successively brought in contact with the two spheres. The new force of repulsion between

\mathrm{P}

and

\mathrm{S}

is :

A 1 N

B 6 N

C 12 N

D 4 N

Correct Answer

Option B

Solution

\begin{aligned} & F_1=\frac{K Q^2}{r^2}=16 \mathrm{~N} \\ & F_2=\frac{K\left(\frac{Q}{2}\right)\left(\frac{3}{4}\right)}{r^2}=\frac{3}{8} \times 16=6 \mathrm{~N} \end{aligned}

Final charges on spheres are

\frac{Q}{2}

and

\frac{3 Q}{4}

.