Electrostatics — Page 4 | JEE Physics

Q31

A solid ball of radius R has a charge density

\rho

given by

\rho

=

\rho

o (1

-

{{r}/{R}}

) for 0

\le

r

\le

R. The electric field outside the ball is :

A

{{{\rho _o}{R^3}} \over {{ \in _o}{r^2}}}

B

{{{\rho _o}{R^3}} \over {12{ \in _o}{r^2}}}

C

{{4{\rho _o}{R^3}} \over {3{ \in _o}{r^2}}}

D

{{3{\rho _o}{R^3}} \over {4{ \in _o}{r^2}}}

Correct Answer

Option B

Solution

Electric field outside the ball is given by E =

{1 \over {4\pi {\varepsilon _0}}}{q \over {{r^2}}}

.............(i) Now, dq = $\rho$ dV = $\rho$ (4 $\pi$ r2)dr $\therefore$ q =

\int {dq = \int\limits_0^R {{\rho _0}\left( {1 - {r \over R}} \right)\left( {4\pi {r^2}} \right)dr} }

=

{\left( {4\pi {\rho _0}} \right)\left[ {{{{r^3}} \over 3} - {1 \over R} \times {{{r^4}} \over 4}} \right]_0^R}

=

{\left( {4\pi {\rho _0}} \right)\left( {{{{R^3}} \over 3} - {{{R^3}} \over 4}} \right)}

$\Rightarrow$ q =

{\left( {4\pi {\rho _0}} \right)\left( {{{{R^3}} \over {12}}} \right)}

......(ii) From eqns. (i) and (ii), E =

{{{\rho _o}{R^3}} \over {12{ \in _o}{r^2}}}

Q32

A body of mass

M

and charge

q

is connected to spring of spring constant

k.

It is oscillating along

x

-direction about its equilibrium position, taken to be at

x=0,

with an amplitude

A

. An electric field

E

is applied along the

x

-direction. Which of the following statements is correct ?

A The new equilibrium position is at a distance

{{qE} \over {2k}}

from

x=0.

B The total energy of the system is

{1 \over 2}m{\omega ^2}{A^2} + {1 \over 2}{{{q^2}{E^2}} \over k}.

C The total energy of the system is

{1 \over 2}m{\omega ^2}{A^2} - {1 \over 2}{{{q^2}{E^2}} \over k}.

D The new equilibrium position is at a distance

{{2qE} \over k}

from

x=0.

Correct Answer

Option B

Solution

On the body of charge q a electric fied E is applied, because of this equilibrium position of body will shift to a point where resulttant force is zero.

\therefore\,\,\,\,

kxeq = qE $\Rightarrow$

\,\,\,

xeq =

{{qE} \over K}

\therefore\,\,\,\,

Total energy of the system =

{1 \over 2}m{\omega ^2}{A^2} + {1 \over 2}K\,x_{eq}^2

=

{1 \over 2}

m

{\omega ^2}

A2 +

{1 \over 2}.{{{q^2}{E^2}} \over K}

Q33

Three concentric metal shells A, B and C of respective radii a, b and c (a < b < c) have surface charge densities

+ \sigma

,

- \sigma

and

+ \sigma

respectively. The potential of shell B is :

A

{\sigma \over { \in {}_0}}\left[ {{{{b^2} - {c^2}} \over c} + a} \right]

B

{\sigma \over { \in {}_0}}\left[ {{{{a^2} - {b^2}} \over a} + c} \right]

C

{\sigma \over { \in {}_0}}\left[ {{{{a^2} - {b^2}} \over b} + c} \right]

D

{\sigma \over { \in {}_0}}\left[ {{{{b^2} - {c^2}} \over b} + a} \right]

Correct Answer

Option C

Solution

Let charge of shell A, B and C are QA, QB and QC respectively.

Potential of B shell will be due to charge QA, QB and QC.

Here charge QA is inside of the shell B and QB is on the surface of the shell B in both cases you have to take the radius of the shell B, while calculating potential of shell B.

Charge QC is outside of the shell B so take radius of shell C for calculating potential of shell B.

$\therefore$ VB = V

_{{Q_a}}

+ V

_{{Q_b}}

+ V

_{{Q_c}}

=

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

\left[ {{{4\pi {a^2}\left( { + \sigma } \right)} \over b} + {{4\pi {b^2}\left( { - \sigma } \right)} \over b} + {{4\pi {c^2}\left( { + \sigma } \right)} \over c}} \right]

=

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

$\times$ 4 $\pi$ $\sigma$

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

=

{\sigma \over {{\varepsilon _0}}}\left[ {{{{a^2} - {b^2}} \over b} + c} \right]

Q34

Let a total charge 2Q be distributed in a sphere of radius R, with the charge density given by

\rho

(r) = kr, where r is the distance from the centre. Two charges A and B, of –Q each, are placed on diametrically opposite points, at equal distance,

a

from the centre. If A and B do not experience any force, then :

A

a = {8^{ - 1/4}}R

B

a = {2^{ - 1/4}}R

C

a = {{3R} \over {{2^{1/4}}}}

D

a = {R \over {\sqrt 3 }}

Correct Answer

Option A

Solution

Total charge = 2Q Charging density $\rho$ = kr Radius = R Charge enclosed in the sphere, qin =

\int\limits_0^R {\rho dV}

$\Rightarrow$ 2Q =

\int\limits_0^R {kr4\pi {r^2}dr}

$\Rightarrow$ 2Q =

k4\pi \int\limits_0^R {{r^3}dr}

$\Rightarrow$ 2Q =

k4\pi {{{R^4}} \over 4}

$\Rightarrow$ k =

{{2Q} \over {\pi {R^4}}}

...........

(1) Force on charge at A will be due to charge at B and due to force applied by the charge in sphere.

Here Fsphere = EQ Using Gauss law, we can find electric field at point A due to sphere, ∮

\overrightarrow E .d\overrightarrow A

=

{{{q_{in}}} \over {{ \in _0}}}

$\Rightarrow$

E\left( {4\pi {a^2}} \right)

=

{{\int\limits_0^a {\rho dV} } \over {{ \in _0}}}

$\Rightarrow$

E\left( {4\pi {a^2}} \right)

=

{{k4\pi {{{a^4}} \over 4}} \over {{ \in _0}}}

$\Rightarrow$ E =

{{k{a^2}} \over {4{ \in _0}}}

As on charge A net force is zero then, FAB = Fsphere $\Rightarrow$

{{Q \times Q} \over {4\pi { \in _0}{{\left( {2a} \right)}^2}}}

=

{{k{a^2}} \over {4{ \in _0}}}

$\times$ Q $\Rightarrow$

{Q \over {4\pi {a^2}}} = k{a^2}

$\Rightarrow$

{Q \over {4\pi {a^2}}} = {{2Q} \over {\pi {R^4}}}{a^2}

[ from equation (1)] $\Rightarrow$

8{a^4} = {R^4}

$\Rightarrow$

a = {8^{ - 1/4}}R

Q35

A point dipole

\overrightarrow p = - {p_0}\widehat x

is kept at the origin. The potential and electric field due to this dipole on the y-axis at a distance d are, respectively: (Take V= 0 at infinity)

A

{{\left| {\overrightarrow p } \right|} \over {4\pi { \in _0}{d^2}}},{{ - \overrightarrow p } \over {4\pi { \in _0}{d^3}}}

B

0,{{\overrightarrow p } \over {4\pi { \in _0}{d^3}}}

C

{{\left| {\overrightarrow p } \right|} \over {4\pi { \in _0}{d^2}}},{{\overrightarrow p } \over {4\pi { \in _0}{d^3}}}

D

0,{{ - \overrightarrow p } \over {4\pi { \in _0}{d^3}}}

Correct Answer

Option D

Solution

V = 0

E = - {{K\overrightarrow P } \over {{r^3}}}

= - {{\overrightarrow p } \over {4\pi {\varepsilon _0}{d^3}}}

Q36

In free space, a particle A of charge 1

\mu

C is held fixed at a point P. Another particle B of the same charge and mass 4

\mu

g is kept at a distance of 1 mm from P. If B is released, then its velocity at a distance of 9 mm from P is :

\left[ {Take\,{1 \over {4\pi { \in _0}}} = 9 \times {{10}^9}N{m^2}{C^{ - 2}}} \right]

A 1.0 m/s

B 6.32

\times

104 m/s

C 2.0

\times

103 m/s

D 1.5

\times

102 m/s

Correct Answer

Option B

Solution

qA = 1 $\mu$ c ; qB = 1 $\mu$ c, mB = 4 × 10–9 kg, rAB = 10–3 m

{1 \over 2}{M_B}{V^2} = k{q_A}{q_B}\left\{ {{1 \over {{{10}^{ - 13}}}} - {1 \over {9 \times {{10}^{ - 3}}}}} \right\}

{1 \over 2}4 \times {10^{ - 9}}{V^2} = 9 \times {10^9} \times {10^{ - 6}} \times {8 \over 9} \times {10^3}

{V^2} = {8 \over 2} \times {10^9} = 4 \times {10^9}

V =

\sqrt {40} \times {10^4}\,m/s

= 6.32 $\times$ 104 m/s

Q37

Two point charges Q each are placed at a distance d apart. A third point charge q is placed at a distance x from mid-point on the perpendicular bisector. The value of x at which charge q will experience the maximum Coulomb's force is :

A x = d

B

x = {d \over 2}

C

x = {d \over {\sqrt 2 }}

D

x = {d \over {2\sqrt 2 }}

Correct Answer

Option D

Solution

Force experienced by the charge q

F = {{kQqx} \over {{{\left[ {{{\left( {{d \over 2}} \right)}^2} + {x^2}} \right]}^{{3 \over 2}}}}}

For maximum Coulomb's force for x

{{dF} \over {dx}} = 0

On solving

x = {d \over {2\sqrt 2 }}

Q38

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}}}

Q39

Four point charges –q, +q, +q and –q are placed on y-axis at y = –2d, y = –d, y = +d and y = +2d, respectively. The magnitude of the electric field E at a point on the x-axis at x = D, with D >> d, will behave as :-

A

E \propto {1 \over D^3}

B

E \propto {1 \over D}

C

E \propto {1 \over D^4}

D

E \propto {1 \over D^2}

Correct Answer

Option C

Solution

Electric field at p = 2E1cos $\theta$ 1 –2E2cos $\theta$ 2 =

{{2Kq} \over {\left( {{d^2} + {D^2}} \right)}} \times {D \over {{{\left( {{d^2} + {D^2}} \right)}^{1/2}}}} - {{2Kq} \over {\left[ {{{\left( {2d} \right)}^2} + {D^2}} \right]}} \times {D \over {{{\left[ {{{\left( {2d} \right)}^2} + {D^2}} \right]}^{1/2}}}}

= 2KqD\left[ {{{\left( {{d^2} + {D^2}} \right)}^{ - 3/2}} - {{\left( {4{d^2} + {D^2}} \right)}^{ - 3/2}}} \right]

= {{2KqD} \over {{D^3}}}\left[ {{{\left( {1 + {{{d^2}} \over {{D^2}}}} \right)}^{ - 3/2}} - {{\left( {1 + {{4{d^2}} \over {{D^2}}}} \right)}^{ - 3/2}}} \right]

Applying binomial approximation $\because$ d << D

= {{2KqD} \over {{D^3}}}\left[ {1 - {3 \over 2}{{{d^2}} \over {{D^2}}} - \left( {1 - {{3 \times 4{d^2}} \over {2{D^2}}}} \right)} \right]

= {{2KqD} \over {{D^3}}}\left[ {{{12} \over 2}{{{d^2}} \over {{D^2}}} - {3 \over 2}{{{d^2}} \over {{D^2}}}} \right]

= {{9kq{d^2}} \over {{D^4}}}

Q40

The electric field in a region is given by

\mathop E\limits^ \to = \left( {Ax + B} \right)\mathop i\limits^ \wedge

, where E is in NC–1 and x is in metres. The values of constants are A = 20 SI unit and B = 10 SI unit. If the potential at x = 1 is V1 and that at x = –5 is V2, then V1 – V2 is :-

A –520 V

B 180 V

C –48 V

D 320 V

Correct Answer

Option B

Solution

\overrightarrow E = (20x + 10)\widehat i

{V_1} - {V_2} = - \int\limits_{ - 5}^1 {\left( {20x + 10} \right)dx}

{V_1} - {V_2} = - \left( {10{x^2} + 10x} \right)_{ - 5}^1

{V_1} - {V_2} = 10\left( {25 - 5 - 1 - 1} \right)

{V_1} - {V_2} = 180\,V