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Electrostatics

NEET Physics · 98 questions · Page 2 of 10 · Click an option or "Show Solution" to reveal answer

Q11
An electric dipole is placed as shown in the figure. The electric potential (in 10 2 V) at point P due to the dipole is (∈0\in_0 = permittivity of free space and 14πϵ0\dfrac{1}{4 \pi \epsilon_{0}} = K) :
A (58)qK\left(\dfrac{5}{8}\right) \mathrm{qK}
B (85)qK\left(\dfrac{8}{5}\right) \mathrm{qK}
C (83)qK\left(\dfrac{8}{3}\right) \mathrm{qK}
D (38)qK\left(\dfrac{3}{8}\right) \mathrm{qK}
Correct Answer
Option D
Solution
v=Kq2×10−2−Kq8×10−2\mathrm{v}=\frac{\mathrm{Kq}}{2 \times 10^{-2}}-\frac{\mathrm{Kq}}{8 \times 10^{-2}}
=Kq[38]×102=\mathrm{Kq}\left[\frac{3}{8}\right] \times 10^{2}
Q12
Six charges +q, −-q, +q, −-q, +q, and −-q are fixed at the corners of a hexagon of side d as shown in the figure. The work done in bringing a charge q 0 to the centre of the hexagon from infinity is (ε0{\varepsilon _0} - permittivity of free space)
A −q24πε0d(6−12){{ - {q^2}} \over {4\pi {\varepsilon _0}d}}\left( {6 - {1 \over {\sqrt 2 }}} \right)
B Zero
C −q24πε0d{{ - {q^2}} \over {4\pi {\varepsilon _0}d}}
D −q24πε0d(3−12){{ - {q^2}} \over {4\pi {\varepsilon _0}d}}\left( {3 - {1 \over {\sqrt 2 }}} \right)
Correct Answer
Option B
Solution

Work done

=q0 . (V0−V∞)= {q_0}\,.\,({V_0} - {V_\infty })
={3kqd+(−3kqd)}q0= \left\{ {{{3kq} \over d} + \left( {{{ - 3kq} \over d}} \right)} \right\}{q_0}

= Zero

Q13
The angle between the electric lines of force and the equipotential surface is
A 0∘^\circ
B 45∘^\circ
C 90∘^\circ
D 180∘^\circ
Correct Answer
Option C
Solution

dV = −-

E→\overrightarrow E

. d

r→\overrightarrow r

dV = −-Edr cosθ\theta For equipotential surface, dV = 0 cosθ\theta = 0 ⇒\Rightarrow θ\theta = 90

∘^\circ
Q14
Two hollow conducting spheres of radii R 1 and R 2 (R 1 >> R 2 ) have equal charges. The potential would be
A More on bigger sphere
B More on smaller sphere
C Equal on both the spheres
D Dependent on the material property of the sphere
Correct Answer
Option B
Solution

Potential of conducting hollow sphere

=KQR= {{KQ} \over R}

Now, Q = same

⇒V∝1R⇒\Rightarrow V \propto {1 \over R} \Rightarrow

more the radius less will be the potential. ⇒\Rightarrow Hence potential would be more on smaller sphere.

Q15
Two point charges −-q and +q are placed at a distance of L, as shown in the figure. The magnitude of electric field intensity at a distance R(R >> L) varies as:
A 1R2{1 \over {{R^2}}}
B 1R3{1 \over {{R^3}}}
C 1R4{1 \over {{R^4}}}
D 1R6{1 \over {{R^6}}}
Correct Answer
Option B
Solution

For R >> L, arrangement is an electric dipole

E=2p4πε0R3E = {{2p} \over {4\pi {\varepsilon _0}{R^3}}}

; where

p=qLp = qL
E∝1R3E \propto {1 \over {{R_3}}}
Q16
Polar molecules are the molecules :
A having a permanent electric dipole moment
B having zero dipole moment
C acquire a dipole moment only in the presence of electric field due to displacement of charges
D acquire a dipole moment only when magnetic field is absent
Correct Answer
Option A
Solution

Polar molecules have centres of positive and negative charges separated by some distance, so they have permanent dipole moment.

Q17
A dipole is placed in an electric field as shown. In which direction will it move?
A towards the right as its potential energy will increase.
B towards the left as its potential energy will increase.
C towards the right as its potential energy will decrease.
D towards the left as its potential energy will decrease.
Correct Answer
Option C
Solution
∣E→1∣>∣E→2∣\left| {{{\overrightarrow E }_1}} \right| > \left| {{{\overrightarrow E }_2}} \right|

As field lines are closed at charge +q.

So, net force on the dipole acts towards right side.

A system always moves to decrease it's potential energy.

Q18
Two charged spherical conductors of radius R 1 and R 2 are connected by a wire. Then the ratio of surface charge densities of the spheres (σ\sigma 1 / σ\sigma 2 ) is :
A R12R22{{R_1^2} \over {R_2^2}}
B R1R2{{R_1^{}} \over {R_2^{}}}
C R2R1{{R_2^{}} \over {R_1^{}}}
D (R1R2)\sqrt {\left( {{{{R_1}} \over {{R_2}}}} \right)}
Correct Answer
Option C
Solution
Q1=∑QR1+R2×R1{Q_1} = {{\sum Q } \over {{R_1} + {R_2}}} \times {R_1}
Q2=∑QR1+R2×R2{Q_2} = {{\sum Q } \over {{R_1} + {R_2}}} \times {R_2}
σ1=Q14πR12=∑QR1+R2×R14πR12∝1R1{\sigma _1} = {{{Q_1}} \over {4\pi R_1^2}} = {{\sum Q } \over {{R_1} + {R_2}}} \times {{{R_1}} \over {4\pi R_1^2}} \propto {1 \over {{R_1}}}
σ2=Q24πR22=∑QR1+R2×R24πR22∝1R2{\sigma _2} = {{{Q_2}} \over {4\pi R_2^2}} = {{\sum Q } \over {{R_1} + {R_2}}} \times {{{R_2}} \over {4\pi R_2^2}} \propto {1 \over {{R_2}}}
σ1σ2=R2R1{{{\sigma _1}} \over {{\sigma _2}}} = {{{R_2}} \over {{R_1}}}
Q19
Twenty seven drops of same size are charged at 220V each. They combine to form a bigger drop. Calculate the potential of the bigger drop.
A 1980 V
B 660 V
C 1320 V
D 1520 V
Correct Answer
Option A
Solution

Let charge and radius of smaller drop is q and r respectively For smaller drop, V =

kqr=220 V{{kq} \over r} = 220\,V

Let R be radius of bigger drop, As volume remains the same

(43πr3)×27=43πR3\left( {{4 \over 3}\pi {r^3}} \right) \times 27 = {4 \over 3}\pi {R^3}

⇒\Rightarrow R = 3r Now, using charge conservation, Q = 27q V big drop =

kQr=k(27q)r=9(kqr){{kQ} \over r} = {{k\left( {27q} \right)} \over r} = 9\left( {{{kq} \over r}} \right)

= 9 ×\times 220 = 1980 V

Q20
A short electric dipole has a dipole moment of 16 × \times 10 -9 Cm. The electric potential due to the dipole at a point at a distance of 0.6 m from the centre of the dipole, situated on a line making an angle of 60∘60^\circ with the dipole axis is : (14πε0=9×109Nm2/C2)\left( {{1 \over {4\pi {\varepsilon _0}}} = 9 \times {{10}^9}N{m^2}/{C^2}} \right)
A 200 V
B 400 V
C zero
D 50 V
Correct Answer
Option A
Solution

Electric potential due to electric dipole (V)

=14πε0p.cos⁡θr2= {1 \over {4\pi {\varepsilon _0}}}{{p.\cos \theta } \over {{r^2}}}
V=9×109×16×109×cos⁡600.36V = {{9 \times {{10}^9} \times 16 \times {{10}^9} \times \cos 60} \over {0.36}}

V = 200 V

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