Electrostatics — Page 8 | NEET Physics

Q71

If the electric flux entering and leaving an enclosed surface respectively is

{\phi _1}

and

{\phi _2},

the electric charge inside the surface will be

A

\left( {{\phi _2} - {\phi _1}} \right){\varepsilon _0}

B

\left( {{\phi _2} + {\phi _1}} \right)/{\varepsilon _0}

C

\left( {{\phi _1} - {\phi _2}} \right)/{\varepsilon _0}

D

\left( {{\phi _1} + {\phi _2}} \right){\varepsilon _0}

Correct Answer

Option A

Solution

The flux entering an enclosed surface is taken as negative and the flux leaving the surface is taken as positive, by convention.

Therefore the net flux leaving the enclosed surface

= {\phi _2} - {\phi _1}

$\therefore$ the change enclosed in the surface by Gauss's law is

q = { \varepsilon _0}\,\left( {{\phi _2} - {\phi _1}} \right)

Q72

A charge particle

'q'

is shot towards another charged particle

'Q'

which is fixed, with a speed

'v'

. It approaches

'Q'

upto a closest distance

r

and then returns. If

q

were given a speed of

'2v'

the closest distances of approaches would be

A

r/2

B

2r

C

r

D

r/4

Correct Answer

Option D

Solution

{1 \over 2}m{v^2} = {{kQq} \over r}

\Rightarrow {1 \over 2}m{\left( {2v} \right)^2} = {{kqQ} \over {r'}}

\Rightarrow r' = {r \over 4}

Q73

A charged oil drop is suspended in a uniform field of

3 \times {10^4}

v/m

so that it neither falls nor rises. The charge on the drop will be (Take the mass of the charge

= 9.9 \times {10^{ - 15}}\,\,kg

and

g = 10\,m/{s^2}

)

A

1.6 \times {10^{ - 18}}\,C

B

3.2 \times {10^{ - 18}}\,C

C

3.3 \times {10^{ - 18}}\,C

D

4.8 \times {10^{ - 18}}\,C

Correct Answer

Option C

Solution

At equilibrium, electric force on drop balances weight of drop.

qE = mg \Rightarrow q

= {{mg} \over E} = {{9.9 \times {{10}^{ - 15}} \times 10} \over {3 \times {{10}^4}}}

= 3.3 \times {10^{ - 18}}C

Q74

(This question contains Statement-1 and Statement-2. Of the four choices given after the statements, choose the one that best describes the two statements.) Statement-1 : For a charged particle moving from point

P

to point

Q

, the net work done by an electrostatic field on the particle is independent of the path connecting point

P

to point

Q.

Statement-2 : The net work done by a conservative force on an object moving along a closed loop is zero.

A Statement-1 is true, Statement-2 is true; Statement-2 is the correct explanation of Statement-1.

B Statement-1 is true, Statement-2 is true; Statement-2 is not the correct explanation of Statement-1.

C Statement- 1 is false, Statement- 2 is true.

D Statement- 1 true, Statement- 2 is false

Correct Answer

Option A

Solution

Statement

1

is true. Statement

2

is true and is the correct explanation of

(1)

Q75

Two identical conducting spheres A and B, carry equal charge. They are separated by a distance much larger than their diameters, and the force between theis F. A third identical conducting sphere, C, is uncharged. Sphere C is first touhed to A, then to B, and then removed. As a result, the force between A and B would be equal to :

A F

B

{{3F} \over 4}

C

{{3F} \over 8}

D

{{F} \over 2}

Correct Answer

Option C

Solution

Let, change of A and B = q

\therefore\,\,\,

Force between them, F =

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

When C touched with A then charge of A. Will fl;ow to C and divide into half parts.

\therefore\,\,\,

charge of A and C , qA = qB =

{q \over 2}

Then C touched with B, then charge on B, qB =

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

\therefore\,\,\,

Force between A and B, F' =

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

=

{{k \times 3{q^2}} \over {8{r^2}}}

=

{3 \over 8} \times {{k{q^2}} \over {{r^2}}}

=

{3 \over 8}\,F

Q76

A uniformly charged ring of radius 3a and total charge q is placed in xy-plane centred at origin. A point charge q is moving towards the ring along the z-axis and has speed u at z = 4a. The minimum value of u such that it crosses the origin is :

A

\sqrt {{2 \over m}} {\left( {{2 \over {15}}{{{q^2}} \over {4\pi {\varepsilon _0}a}}} \right)^{1/2}}

B

\sqrt {{2 \over m}} {\left( {{1 \over {15}}{{{q^2}} \over {4\pi {\varepsilon _0}a}}} \right)^{1/2}}

C

\sqrt {{2 \over m}} {\left( {{1 \over {5}}{{{q^2}} \over {4\pi {\varepsilon _0}a}}} \right)^{1/2}}

D

\sqrt {{2 \over m}} {\left( {{4 \over {15}}{{{q^2}} \over {4\pi {\varepsilon _0}a}}} \right)^{1/2}}

Correct Answer

Option A

Solution

Ui + Ki = Uf + Kf

{{k{q^2}} \over {\sqrt {16{a^2} + 9{a^2}} }} + {1 \over 2}m{v^2} = {{k{q^2}} \over {3a}}

{1 \over 2}m{v^2} = {{k{q^2}} \over a}\left( {{1 \over 3} - {1 \over 5}} \right) = {{2k{q^2}} \over {15a}}

v = \sqrt {{{4k{q^2}} \over {15ma}}}

$\therefore$ v =

\sqrt {{2 \over m}} {\left( {{{2{q^2}} \over {15 \times 4\pi {\varepsilon _0}a}}} \right)^{{1 \over 2}}}

Q77

Ten charges are placed on the circumference of a circle of radius R with constant angular separation between successive charges. Alternate charges 1, 3, 5, 7, 9 have charge (+q) each, while 2, 4, 6, 8, 10 have charge (–q) each. The potential V and the electric field E at the centre of the circle are respectively. (Take V = 0 at infinity)

A V = 0; E = 0

B

V = {{10q} \over {4\pi {\varepsilon _0}R}}

;

E = {{10q} \over {4\pi {\varepsilon _0}{R^2}}}

C

V = {{10q} \over {4\pi {\varepsilon _0}R}}

; E = 0

D V = 0;

E = {{10q} \over {4\pi {\varepsilon _0}{R^2}}}

Correct Answer

Option A

Solution

Net charge = 5q - 5q = 0 Potential of centre = V =

{{K\sum q } \over r}

VC =

{{K\left( 0 \right)} \over r}

= 0 Let E be electric field produced by each charge at the centre, then resultant electric field will be EC = 0, Since equal electric field vectors are acting at equal angle so their resultant is equal to zero.(

From symmetric property of vector)

Q78

Find out the surface charge density at the intersection of point x = 3 m plane and x-axis, in the region of uniform line charge of 8 nC/m lying along the z-axis in free space.

A 0.424 nC m

-

2

B 4.0 nC m

-

2

C 47.88 C/m

D 0.07 nC m

-

2

Correct Answer

Option A

Solution

Electric field due to wire is given by

E = {{2k\lambda } \over r}

Electric field with surface charge density

E = {\sigma \over {{ \in _0}}}

{{2k\lambda } \over r} = {\sigma \over {{ \in _0}}}

\Rightarrow 2{1 \over {4\pi { \in _0}}}{\lambda \over r} = {\sigma \over {{ \in _0}}}

\Rightarrow {8 \over {2 \times 3.14 \times 3}} = \sigma

\Rightarrow \sigma = 0.424

n Cm $-$ 2

Q79

A long cylindrical volume contains a uniformly distributed charge of density

\rho

. The radius of cylindrical volume is R. A charge particle (q) revolves around the cylinder in a circular path. The kinetic energy of the particle is :

A

{{\rho q{R^2}} \over {4{\varepsilon _0}}}

B

{{\rho q{R^2}} \over {2{\varepsilon _0}}}

C

{{q\rho } \over {4{\varepsilon _0}{R^2}}}

D

{{4{\varepsilon _0}{R^2}} \over {q\rho }}

Correct Answer

Option A

Solution

{{m{v^2}} \over r} = {{2k\rho \times \pi {R^2}q} \over r}

\Rightarrow {1 \over 2}m{v^2} = {{\rho {R^2}q} \over {4{\varepsilon _0}}}

Q80

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.