Electrostatics — Page 12 | JEE Physics

Q111

Within a spherical charge distribution of charge density

\rho

(r), N equipotential surfaces of potential V0, V0 +

\Delta

V, V0 + 2

\Delta

V, .......... V0 + N

\Delta

V (

\Delta

V > 0), are drawn and have increasing radii r0, r1, r2,..........rN, respectively. If the difference in the radii of the surfaces is constant for all values of V0 and

\Delta

V then :

A

\rho

(r)

\alpha

r

B

\rho

(r) = constant

C

\rho

(r)

\alpha

{1 \over r}

D

\rho

(r)

\alpha

{1 \over {{r^2}}}

Correct Answer

Option C

Solution

Here,

\Delta

v and

\Delta

r are same for any pair of surface. we know, Electric field, E = $-$

{{dv} \over {dr}}

$\therefore$ E = constant [As dv and dr are constant] Electric field inside the spherical charge distribution. E =

{{\rho r} \over {3{\varepsilon _0}}}

Now, as E = constant $\therefore$ $\rho$ r = constant $\Rightarrow$ $\rho$ (r) $\propto$

{1 \over r}

Q112

Four equal point charges Q each are placed in the xy plane at (0, 2), (4, 2), (4, –2) and (0, –2). The work required to put a fifth charge Q at the origin of the coordinate system will be -

A

{{{Q_2}} \over {4\pi {\varepsilon _0}}}

B

{{{Q^2}} \over {2\sqrt 2 \pi {\varepsilon _0}}}

C

{{{Q_2}} \over {4\pi {\varepsilon _0}}}\left( {1 + {1 \over {\sqrt 3 }}} \right)

D

{{{Q_2}} \over {4\pi {\varepsilon _0}}}\left( {1 + {1 \over {\sqrt 5 }}} \right)

Correct Answer

Option D

Solution

Potential at origin =

{{KQ} \over 2} + {{KQ} \over 2} + {{KQ} \over {\sqrt {20} }} + {{KQ} \over {\sqrt {20} }}

(Potential at $\infty$ = 0) = KQ

\left( {1 + {1 \over {\sqrt 5 }}} \right)

$\therefore$ Work required to put a fifth charge Q at origin is equal to

{{{Q^2}} \over {4\pi {\varepsilon _0}}}\left( {1 + {1 \over {\sqrt 5 }}} \right)

Q113

Charge is distributed within a sphere of radius R with a volume charge density

\rho \left( r \right) = {A \over {{r^2}}}{e^{ - {{2r} \over s}}},

where A and a are constants. If Q is the total charge of this charge distribution, the radius R is :

A a log

\left( {1 - {Q \over {2\pi aA}}} \right)

B

{a \over 2}

log

\left( {{1 \over {1 - {Q \over {2\pi aA}}}}} \right)

C a log

\left( {{1 \over {1 - {Q \over {2\pi aA}}}}} \right)

D

{a \over 2}

log

\left( {1 - {Q \over {2\pi aA}}} \right)

Correct Answer

Option B

Solution

Volume of this spherical layer, dv = (4 $\pi$ r2)dr charge present in this layer, dq = $\rho$ (4 $\pi$ r2 dr) =

{A \over {{r^2}}}{e^{ - {{2r} \over a}}}\,\,\left( {4\pi {r^2}dr} \right)

=

A\,{e^{ - {{2r} \over a}}}\left( {4\pi dr} \right)

$\therefore$ Total charge in the sphere, Q=

\int\limits_0^R {4\pi A{e^{ - {{2r} \over a}}}} \,dr

= 4 $\pi$ A

\int\limits_0^R {{e^{ - {{2r} \over a}}}} \,dr

= 4 $\pi$ A

\left[ {{{{e^{ - {{2r} \over a}}}} \over { - {2 \over a}}}} \right]_0^R

= 4 $\pi$ A

\left( { - {a \over 2}} \right)\left( {{e^{ - {{2R} \over a}}} - 1} \right)

$\therefore$ Q = 2 $\pi$ aA

\left( {1 - {e^{ - {{2R} \over a}}}} \right)

$\Rightarrow$

{1 - {e^{ - {{2R} \over a}}}}

=

{Q \over {2\pi aA}}

$\Rightarrow$

{{e^{ - {{2R} \over a}}}}

= 1 $-$

{Q \over {2\pi aA}}

$\Rightarrow$

{e^{{{2R} \over a}}}

=

{1 \over {1 - {Q \over {2\pi aA}}}}

$\Rightarrow$

{{2R} \over a} = \log \left( {{1 \over {1 - {Q \over {2\pi aA}}}}} \right)

$\Rightarrow$ R =

{a \over 2}

log

\left( {{1 \over {1 - {Q \over {2\pi aA}}}}} \right)

Q114

A charge Q is distributed over three concentric spherical shells of radii a, b, c (a < b < c) such that their surface charge densities are equal to one another. The total potential at a point at distance r from their common centre, where r < a, would be -

A

{{Q\left( {{a^2} + {b^2} + {c^2}} \right)} \over {4\pi {\varepsilon _0}\left( {{a^3} + {b^3} + {c^3}} \right)}}

B

{Q \over {4\pi {\varepsilon _0}\left( {a + b + c} \right)}}

C

{Q \over {12\pi {\varepsilon _0}}}{{ab + bc + ca} \over {abc}}

D

{{Q\left( {a + b + c} \right)} \over {4\pi {\varepsilon _0}\left( {{a^2} + {b^2} + {c^2}} \right)}}

Correct Answer

Option D

Solution

Potential at point P, V =

{{k{Q_a}} \over a} + {{k{Q_b}} \over b} + {{k{Q_c}} \over c}

$\because$ Qa : Qb : Qc : : a2 : b2 : c2 {since $\sigma$ a = $\sigma$ b = $\sigma$ c} $\therefore$ Qa =

\left[ {{{{a^2}} \over {{a^2} + {b^2} + {c^2}}}} \right]

Q Qb =

\left[ {{{{b^2}} \over {{a^2} + {b^2} + {c^2}}}} \right]

Q Qc =

\left[ {{{{c^2}} \over {{a^2} + {b^2} + {c^2}}}} \right]

Q V =

{Q \over {4\pi { \in _0}}}\left[ {{{\left( {a + b + c} \right)} \over {{a^2} + {b^2} + {c^2}}}} \right]

Q115

Two charges of

5 Q

and

-2 Q

are situated at the points

(3 a, 0)

and

(-5 a, 0)

respectively. The electric flux through a sphere of radius '

4 a

' having center at origin is :

A

\dfrac{2 Q}{\varepsilon_0}

B

\dfrac{7 Q}{\varepsilon_0}

C

\dfrac{3 Q}{\varepsilon_0}

D

\dfrac{5 Q}{\varepsilon_0}

Correct Answer

Option D

Solution

The electric flux through any closed surface is given by Gauss's law, which can be stated as:

\Phi = \frac{Q_{\text{enc}}}{\varepsilon_0}

where:

\Phi

= electric flux

Q_{\text{enc}}

= total charge enclosed by the surface

\varepsilon_0

= permittivity of free space (electric constant) In this scenario, we have a sphere of radius

4a

with its center at the origin and two charges,

5Q

at point (3a, 0) and

-2Q

at point (-5a, 0). Since the sphere's radius is

4a

, the charge

5Q

, which is located at (3a, 0), lies inside the sphere, whereas the charge

-2Q

, located at (-5a, 0), lies outside the sphere.

Gauss's law only considers charges that are enclosed within the surface.

Thus, the charge

-2Q

has no impact on the electric flux through the sphere because it is not enclosed by the sphere. Only the charge

5Q

contributes to the electric flux inside the sphere.

The electric flux through the sphere is therefore given simply by the enclosed charge:

\Phi = \frac{5Q}{\varepsilon_0}

Q116

If

\epsilon_0

denotes the permittivity of free space and

\Phi_E

is the flux of the electric field through the area bounded by the closed surface, then dimensions of

\left(\epsilon_0 \dfrac{d \phi_E}{d t}\right)

are that of :

A electric charge

B electric field

C electric current

D electric potential

Correct Answer

Option C

Solution

We can use Gauss’s law to analyze the dimensions. Gauss’s law states that

\Phi_E = \frac{q}{\epsilon_0},

where

\Phi_E

is the electric flux,

q

is the electric charge,

\epsilon_0

is the permittivity of free space. To find the dimensions of

\epsilon_0 \frac{d\Phi_E}{dt}

, follow these steps: Differentiate Gauss’s law with respect to time:

\frac{d\Phi_E}{dt} = \frac{1}{\epsilon_0}\frac{dq}{dt}.

Multiply the equation by

\epsilon_0

:

\epsilon_0 \frac{d\Phi_E}{dt} = \frac{dq}{dt}.

Recognize that the time rate of change of charge,

\frac{dq}{dt}

, is by definition the electric current. Thus, the dimensions of

\epsilon_0 \frac{d\Phi_E}{dt}

are those of electric current. The correct answer is: Option C, electric current.

Q117

Two electrons each are fixed at a distance '2d'. A third charge proton placed at the midpoint is displaced slightly by a distance x (x << d) perpendicular to the line joining the two fixed charges. Proton will execute simple harmonic motion having angular frequency : (m = mass of charged particle)

A

{\left( {{{2{q^2}} \over {\pi {\varepsilon _0}m{d^3}}}} \right)^{{1 \over 2}}}

B

{\left( {{{{q^2}} \over {2\pi {\varepsilon _0}m{d^3}}}} \right)^{{1 \over 2}}}

C

{\left( {{{2\pi {\varepsilon _0}m{d^3}} \over {{q^2}}}} \right)^{{1 \over 2}}}

D

{\left( {{{\pi {\varepsilon _0}m{d^3}} \over {2{q^2}}}} \right)^{{1 \over 2}}}

Correct Answer

Option B

Solution

The arrangement of charges is shown below As we know that, Coulomb's force between two charges. i.e., q1 and q2,

F = {1 \over {4\pi {\varepsilon _0}}}{{{q_1}{q_2}} \over {{r^2}}} = {1 \over {4\pi {\varepsilon _0}}}{{{q_1}{q_2}} \over {({d^2} + {x^2})}}

..... (i) Here,

{q_1} = {q_2} = q

Force in SHM,

F = m{\omega ^2}x

...... (ii) Since, in order to have SHM +q should move downwards and force responsible for this will be only

F' = F\sin \theta + F\sin \theta = 2F\sin \theta

..... (iii) Using Eqs. (ii) and (iii), we get

2F\sin \theta = m{\omega ^2}x

\Rightarrow {2 \over {4\pi {\varepsilon _0}}}{{{q^2}} \over {({d^2} + {x^2})}}\sin \theta = m{\omega ^2}x

\Rightarrow {2 \over {4\pi {\varepsilon _0}}}{{{q^2}} \over {({d^2} + {x^2})}}.{x \over {{{({d^2} + {x^2})}^{1/2}}}} = m{\omega ^2}x

\Rightarrow \omega = {\left( {{1 \over {2\pi {\varepsilon _0}}}{{{q^2}} \over {{{({d^2} + {x^2})}^{3/2}}m}}} \right)^{1/2}}

As,

x < < d

$\therefore$

\omega = {\left( {{1 \over {2\pi {\varepsilon _0}}}{{{q^2}} \over {m{d^3}}}} \right)^{1/2}}

Q118

In hydrogen like system the ratio of coulombian force and gravitational force between an electron and a proton is in the order of :

A 1019

B 1039

C 1029

D 1036

Correct Answer

Option B

Solution

To find the ratio of the Coulombic force to the gravitational force between an electron and a proton in a hydrogen-like system, we use the formulae for both forces and then divide them.

The Coulombic (electrostatic) force, $F_C$ , between two charges is given by Coulomb's law: $F_C = k \dfrac{|q_1 q_2|}{r^2}$ where $k$ is Coulomb's constant ( $8.987 \times 10^9 \, \text{Nm}^2\text{C}^{-2}$ ), $q_1$ and $q_2$ are the magnitudes of the charges, and $r$ is the distance between the charges.

For a proton and an electron, $q_1 = q_2 = e$ , where $e$ is the elementary charge ( $1.602 \times 10^{-19} \, \text{C}$ ).

The gravitational force, $F_G$ , between two masses is given by Newton's law of universal gravitation: $F_G = G \dfrac{m_1 m_2}{r^2}$ where $G$ is the gravitational constant ( $6.674 \times 10^{-11} \, \text{Nm}^2\text{kg}^{-2}$ ), and $m_1$ and $m_2$ are the masses of the two objects.

For a proton and an electron, $m_p \approx 1.673 \times 10^{-27} \, \text{kg}$ and $m_e \approx 9.109 \times 10^{-31} \, \text{kg}$ , respectively.

The ratio of the Coulombic to the gravitational force is therefore: $\dfrac{F_C}{F_G} = \dfrac{k \dfrac{|e^2|}{r^2}}{G \dfrac{m_p m_e}{r^2}} = \dfrac{k e^2}{G m_p m_e}$ Plugging in the values: $\dfrac{F_C}{F_G} = \dfrac{(8.987 \times 10^9) (1.602 \times 10^{-19})^2}{(6.674 \times 10^{-11}) (1.673 \times 10^{-27}) (9.109 \times 10^{-31})}$ $\dfrac{F_C}{F_G} = \dfrac{(8.987 \times 10^9) \times (2.568 \times 10^{-38})}{(6.674 \times 10^{-11}) \times (1.523 \times 10^{-57})}$ $\dfrac{F_C}{F_G} \approx 2.3 \times 10^{39}$ Therefore, the ratio of the Coulombic force to the gravitational force between an electron and a proton in a hydrogen-like system is on the order of $10^{39}$ , making Option B the correct answer.

Q119

A vertical electric field of magnitude 4.9

\times

105 N/C just prevents a water droplet of a mass 0.1 g from falling. The value of charge on the droplet will be : (Given : g = 9.8 m/s2)

A 1.6

\times

10

-

9 C

B 2.0

\times

10

-

9 C

C 3.2

\times

10

-

9 C

D 0.5

\times

10

-

9 C

Correct Answer

Option B

Solution

Since the droplet is at rest $\Rightarrow$ Net force = 0 $\Rightarrow$ mg = qE $\Rightarrow$

q = {{mg} \over E}

= 2 $\times$ 10 $-$ 9 C

Q120

Two identical charged spheres are suspended by strings of equal lengths. The strings make an angle of

{30^ \circ }

with each other. When suspended in a liquid of density

0.8g

c{m^{ - 3}},

the angle remains the same. If density of the material of the sphere is

1.6

g

c{m^{ - 3}},

the dielectric constant of the liquid is

A

4

B

3

C

2

D

1

Correct Answer

Option C

Solution

{F_e} = T\sin {15^ \circ }\,\,;

mg = T\cos {15^ \circ }

\Rightarrow \tan {15^ \circ } = {{{F_e}} \over {mg}}

\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)

In liquid,

{F_e}' = T'\sin {15^ \circ }

\,\,\,\,\,\,...(ii)

mg = {F_B} + T'\cos {15^ \circ }

{F_B}' = V\left( {d - \rho } \right)g = V\left( {1.6 - 0.8} \right)g = 0.8\,Vg

= 0.8{m \over d}g = {{0.8mg} \over {1.6}} = {{mg} \over 2}

$\therefore$

mg = {{mg} \over 2} + T'\cos {15^ \circ }

\Rightarrow {{mg} \over 2} = T'\cos {15^ \circ }

\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( B \right)

From

(A)

and

(B),

\tan \,{15^ \circ } = {{2{F_e}'} \over {mg}}\,\,\,\,\,\,\,\,\,...\left( 2 \right)

From

(1)

and

(2)

{{{F_e}} \over {mg}} = {{2{F_e}'} \over {mg}} \Rightarrow {F_e} = 2{F_e}' \Rightarrow {F_e}' = {{{F_e}} \over 2}