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Electrostatics

JEE Physics · 130 questions · Page 3 of 13 · Click an option or "Show Solution" to reveal answer

Q21
(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 PP to point QQ, the net work done by an electrostatic field on the particle is independent of the path connecting point PP to point Q.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

11

is true. Statement

22

is true and is the correct explanation of

(1)(1)
Q22
Two charges, each equals to q,q, are kept at x=−ax=-a and x=ax=a on the xx-axis. A particle of mass mm and charge q0=q2{q_0} = {q \over 2} is placed at the origin. If charge q0{q_0} is given a small displacement (y<<a)\left( {y < < a} \right) along the yy-axis, the net force acting on the particle is proportional to
A yy
B −y-y
C 1y{1 \over y}
D −1y-{1 \over y}
Correct Answer
Option A
Solution
⇒Fnet=2F cos⁡θ\Rightarrow {F_{net}} = 2F\,\cos \theta
Fnet=2kq(q2)(y2+a2)2.yy2+a2{F_{net}} = {{2kq\left( {{q \over 2}} \right)} \over {{{\left( {\sqrt {{y^2} + {a^2}} } \right)}^2}}}.{y \over {\sqrt {{y^2} + {a^2}} }}
Fnet=2kq(q2)y(y2+a2)3/2⇒kq2ya3{F_{net}} = {{2kq\left( {{q \over 2}} \right)y} \over {{{\left( {{y^2} + {a^2}} \right)}^{3/2}}}} \Rightarrow {{k{q^2}y} \over {{a^3}}}

S0,

F∝yF \propto y
Q23
Let there be a spherically symmetric charge distribution with charge density varying as ρ(r)=ρ0(54−rR)\rho \left( r \right) = {\rho _0}\left( {{5 \over 4} - {r \over R}} \right) upto r=R,r=R, and ρ(r)=0\rho \left( r \right) = 0 for r>R,r>R, where rr is the distance from the erigin. The electric field at a distance r(r<R)r\left( {r < R} \right) from the origin is given by
A ρ0r4ε0(53−rR){{{\rho _0}r} \over {4{\varepsilon _0}}}\left( {{5 \over 3} - {r \over R}} \right)
B 4πρ0r3ε0(53−rR){{4\pi {\rho _0}r} \over {3{\varepsilon _0}}}\left( {{5 \over 3} - {r \over R}} \right)
C 4ρ0r4ε0(54−rR){{4{\rho _0}r} \over {4{\varepsilon _0}}}\left( {{5 \over 4} - {r \over R}} \right)
D ρ0r3ε0(54−rR){{{\rho _0}r} \over {3{\varepsilon _0}}}\left( {{5 \over 4} - {r \over R}} \right)
Correct Answer
Option A
Solution

Let us consider a spherical shell of radius

xx

and thickness

dx.dx.

Charge on this shell

dq=ρ.4π2dx=ρ0(54−xR).4πx2dxdq = \rho .4{\pi ^2}dx = {\rho _0}\left( {{5 \over 4} - {x \over R}} \right).4\pi {x^2}dx

∴\therefore Total charge in the spherical region from center to

rr
(r<R)\left( {r < R} \right)

is

q=∫dq=4πρ0∫0r(54−xR)x2dxq = \int {dq = 4\pi {\rho _0}\int\limits_0^r {\left( {{5 \over 4} - {x \over R}} \right)} } {x^2}dx
=4πρ0[54.r33−1R.r44]= 4\pi {\rho _0}\left[ {{5 \over 4}.{{{r^3}} \over 3} - {1 \over R}.{{{r^4}} \over 4}} \right]
=πρ0r3(53−rR)= \pi {\rho _0}{r^3}\left( {{5 \over 3} - {r \over R}} \right)

∴\therefore Electric field at

r,r,
E=14π∈0.qr2E = {1 \over {4\pi { \in _0}}}.{q \over {{r^2}}}
=14π∈0.πρ0r3r2(53−rR)= {1 \over {4\pi { \in _0}}}.{{\pi {\rho _0}{r^3}} \over {{r^2}}}\left( {{5 \over 3} - {r \over R}} \right)
=ρ0r4∈0(53−rR)= {{{\rho _0}r} \over {4{ \in _0}}}\left( {{5 \over 3} - {r \over R}} \right)
Q24
Two identical charged spheres suspended from a common point by two massless strings of length ll are initially a distance d(d<<1)d\left( {d < < 1} \right) apart because of their mutual repulsion. The charge begins to leak from both the spheres at a constant rate. As a result charges approach each other with a velocity vv. Then as a function of distance xx between them,
A v ∝ x−1v\, \propto \,{x^{ - 1}}
B y ∝ x1/2y\, \propto \,{x^{{{1}/{2}}}}
C v ∝ xv\, \propto \,x
D v ∝ x−1/2v\, \propto \,{x^{ - {{1}/{2}}}}
Correct Answer
Option D
Solution

At any instant

Tcos⁡θ=mg                ...(i)T\cos \theta = mg\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( i \right)
Tsin⁡θ=Fe                ...(ii)T\sin \theta = {F_e}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( {ii} \right)
⇒sin⁡θcos⁡θ=Femg⇒Fe=mg tan⁡θ\Rightarrow {{\sin \theta } \over {\cos \theta }} = {{{F_e}} \over {mg}} \Rightarrow {F_e} = mg\,\tan \theta
⇒kq2x2=mg tan⁡θ⇒q2∝x2tan⁡θ\Rightarrow {{k{q^2}} \over {{x^2}}} = mg\,\tan \theta \Rightarrow {q^2} \propto {x^2}\tan \theta
sin⁡θ=x2l\sin \theta = {\textstyle{x \over {2l}}}

For small

θ, sin⁡θ≈tan⁡θ\theta ,\,\sin \theta \approx \tan \theta

∴\therefore

q2∝x3{q^2} \propto {x^3}
⇒qdqdt∝x2dxdt\Rightarrow q{{dq} \over {dt}} \propto {x^2}{{dx} \over {dt}}

∴\therefore

dqdt=const.{{dq} \over {dt}} = const.

∴\therefore

q∝x2.v⇒x3/2αx2.v  q \propto {x^2}.v \Rightarrow {x^{3/2}}\alpha {x^2}.v\,\,
     \,\,\,\,\,
[  \left[ {\,\,} \right.

as

q2∝x3  ]\left. {{q^2} \propto {x^3}\,\,} \right]
⇒v∝x−1/2\Rightarrow v \propto {x^{ - 1/2}}
Q25
The electrostatic potential inside a charged spherical ball is given by ϕ=ar2+b\phi = a{r^2} + b where rr is the distance from the center and a,ba,b are constants. Then the charge density inside the ball is:
A −6aε0r - 6a{\varepsilon _0}r
B −24πaε0 - 24\pi a{\varepsilon _0}
C −6aε0 - 6a{\varepsilon _0}
D −24πε0r - 24\pi {\varepsilon _0}r
Correct Answer
Option C
Solution

Electric field

E=dϕdr=−2ar          ...(i)E = {{d\phi } \over {dr}} = - 2ar\,\,\,\,\,\,\,\,\,\,...\left( i \right)

By Gauss's theorem

E=14θε0r2               ...(ii)E = {1 \over {4\theta {\varepsilon _0}{r^2}}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( {ii} \right)

From

(i)\left( i \right)

and

(ii),\left( ii \right),
q=−8πε0ar3q = - 8\pi {\varepsilon _0}a{r^3}
⇒dq=−24πε0ar2dr\Rightarrow dq = - 24\pi {\varepsilon _0}ar{}^2dr

Charge density,

ρ=dq4πr2dr=−6ε0a\rho = {{dq} \over {4\pi {r^2}dr}} = - 6{\varepsilon _0}a
Q26
This question has statement- 11 and statement- 2.2. Of the four choices given after the statements, choose the one that best describe the two statements. An insulating solid sphere of radius RR has a uniformly positive charge density ρ\rho . As a result of this uniform charge distribution there is a finite value of electric potential at the center of the sphere, at the surface of the sphere and also at a point out side the sphere. The electric potential at infinite is zero. Statement- 1:1: When a charge qq is take from the centre of the surface of the sphere its potential energy changes by qρ3ε0{{q\rho } \over {3{\varepsilon _0}}} Statement- 2:2: The electric field at a distance r(r<R)r\left( {r < R} \right) from the center of the sphere is ρr3ε0.{{\rho r} \over {3{\varepsilon _0}}}.
A Statement- 11 is true, Statement- 22 is true; Statement- 22 is not the correct explanation of Statement- 11.
B Statement 11 is true, Statement 22 is false.
C Statement 11 is false, Statement 22 is true.
D Statement- 11 is true, Statement- 22 is true; Statement- 22 is the correct explanation of Statement- 11.
Correct Answer
Option C
Solution

The electric field inside a uniformly charged sphere is =

ρ.r3∈0{{\rho .r} \over {3{ \in _0}}}

The electric potential inside a uniformly charged sphere

=ρR26∈0[3−r2R2]= {{\rho {R^2}} \over {6{ \in _0}}}\left[ {3 - {{{r^2}} \over {{R^2}}}} \right]

∴\therefore Potential difference between center and surface

=ρR26∈0[3−2]=ρR26∈0= {{\rho {R^2}} \over {6{ \in _0}}}\left[ {3 - 2} \right] = {{\rho {R^2}} \over {6{ \in _0}}}
ΔU=qρR26∈0\Delta U = {{q\rho {R^2}} \over {6{ \in _0}}}
Q27
Assume that an electric field E→=30x2i^\overrightarrow E = 30{x^2}\widehat i exists in space. Then the potential difference VA−VO,{V_A} - {V_O}, where VO{V_O} is the potential at the origin and VA{V_A} the potential at x=2x=2 mm is :
A 120120 J/CJ/C
B −120-120 J/CJ/C
C −80-80 J/CJ/C
D 8080 J/CJ/C
Correct Answer
Option C
Solution

Potential difference between any two points in an electric field is given by,

dV=−E→.dx→dV = - \overrightarrow E .\overrightarrow {dx}
∫VOVAdV=−∫0230x2dx\int\limits_{{V_O}}^{{V_A}} {dV = - \int\limits_0^2 {30{x^2}} } dx
VA−VO=[10x3]02=−80J/C{V_A} - {V_O} = \left[ {10{x^3}} \right]_{0}^2 = - 80J/C
Q28
An electric dipole has a fixed dipole moment p→\overrightarrow p , which makes angle θ\theta with respect to x-axis. When subjected to an electric field E1→=Ei^\mathop {{E_1}}\limits^ \to = E\widehat i , it experiences a torque T1→=τk^\overrightarrow {{T_1}} = \tau \widehat k . When subjected to another electric field E2→=3E1j^\mathop {{E_2}}\limits^ \to = \sqrt 3 {E_1}\widehat j it experiences a torque T2→=−T1→\mathop {{T_2}}\limits^ \to = \mathop { - {T_1}}\limits^ \to . The angle θ\theta is:
A 90o
B 45o
C 30o
D 60o
Correct Answer
Option D
Solution

Torque experienced by the dipole in an electric field,

TT

= pE sinθ\theta

T→=p→×E→\overrightarrow T = \overrightarrow p \times \overrightarrow E
p→=pcos⁡θi^+psin⁡θj^\overrightarrow p = p\cos \theta \widehat i + p\sin \theta \widehat j
E1→=Ei^\mathop {{E_1}}\limits^ \to = E\widehat i
T1→=P→×E→1\overrightarrow {{T _1}} = \overrightarrow P \times {\overrightarrow E _1}

= (

pcos⁡θi^+psin⁡θj^p\cos \theta \widehat i + p\sin \theta \widehat j

) ×\times

E(i^)E\left( {\widehat i} \right)

= pE sinθ\theta

(−k^)\left( { - \widehat k} \right)
E2→=3E1j^\mathop {{E_2}}\limits^ \to = \sqrt 3 {E_1}\widehat j
T2→=\overrightarrow {{T _2}} =

(

pcos⁡θi^+psin⁡θj^p\cos \theta \widehat i + p\sin \theta \widehat j

) ×\times

3E1j^\sqrt 3 {E_1}\widehat j

=

3pEcos⁡θ(k^)\sqrt 3 pE\cos \theta \left( {\widehat k} \right)

Now given,

T2→\overrightarrow {{T _2}}

=

−T1→-\overrightarrow {{T _1}}

⇒\Rightarrow

3pEcos⁡θ(k^)\sqrt 3 pE\cos \theta \left( {\widehat k} \right)

= -pE sinθ\theta

(−k^)\left( { - \widehat k} \right)

⇒\Rightarrow

tan⁡θ=3\tan \theta = \sqrt 3

⇒\Rightarrow θ\theta = 60o

Q29
There is a uniform electrostatic field in a region. The potential at various points on a small sphere centred at P,P, in the region, is found to vary between the limits 589.0 V to 589.8 V. What is the potential at a point on the sphere whose radius vector makes an angle of 60o with the direction of the field ?
A 589.5 V
B 589.2 V
C 589.4 V
D 589.6 V
Correct Answer
Option C
Solution

Potential gradient,

Δ\Delta

V = E. d ⇒\Rightarrow

   \,\,\,

589.8 −- 589.0 = (E d)max ⇒\Rightarrow

   \,\,\,

(E d)max = 0.8

∴   \therefore\,\,\,
Δ\Delta

V = E d cosθ\theta = 0.8 ×\times cos60o = 0.4

∴   \therefore\,\,\,

Maximum potential on the sphere = 589.4 V

Q30
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 3F4{{3F} \over 4}
C 3F8{{3F} \over 8}
D F2{{F} \over 2}
Correct Answer
Option C
Solution

Let, change of A and B = q

∴   \therefore\,\,\,

Force between them, F =

k×q×qr2=kq2r2{{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 =

q2{q \over 2}

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

q2+q2=3q4{{{q \over 2} + q} \over 2} = {{3q} \over 4}
∴   \therefore\,\,\,

Force between A and B, F' =

k×q2×3q4r2{{k \times {q \over 2} \times {{3q} \over 4}} \over {{r^2}}}

=

k×3q28r2{{k \times 3{q^2}} \over {8{r^2}}}

=

38×kq2r2{3 \over 8} \times {{k{q^2}} \over {{r^2}}}

=

38 F{3 \over 8}\,F
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