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Electrostatics

JEE Physics · 130 questions · Page 6 of 13 · Click an option or "Show Solution" to reveal answer

Q51
In finding the electric field using Gauss Law the formula ∣E→∣=qencε0∣A∣\left| {\overrightarrow E } \right| = {{{q_{enc}}} \over {{\varepsilon _0}\left| A \right|}} is applicable. In the formula ε0{{\varepsilon _0}} is permittivity of free space, A is the area of Gaussian surface and qenc is charge enclosed by the Gaussian surface. The equation can be used in which of the following situation?
A Only when ∣E→∣\left| {\overrightarrow E } \right| = constant on the surface.
B For any choice of Gaussian surface.
C Only when the Gaussian surface is an equipotential surface.
D Only when the Gaussian surface is an equipotential surface and ∣E→∣\left| {\overrightarrow E } \right| is constant on the surface.
Correct Answer
Option D
Solution

By Gauss law

∮E→.dA→=qinε0\oint {\overrightarrow E .d\overrightarrow A } = {{{q_{in}}} \over {{\varepsilon _0}}}

When

E→∣∣A→{\overrightarrow E ||\overrightarrow A }

and

∣E→∣{\left| {\overrightarrow E } \right|}

is constant then

E∫dA=qinε0E\int {dA} = {{{q_{in}}} \over {{\varepsilon _0}}}

⇒\RightarrowEA =

qinε0{{{q_{in}}} \over {{\varepsilon _0}}}

∴\therefore

∣E→∣{\left| {\overrightarrow E } \right|}

should be constant on the surface and the surface should be equipotential.

Q52
An electric dipole of moment p→=(−i^−3j^+2k^)×10−29\overrightarrow p = \left( { - \widehat i - 3\widehat j + 2\widehat k} \right) \times {10^{ - 29}} C.m is at the origin (0, 0, 0). The electric field due to this dipole at r→=+i^+3j^+5k^\overrightarrow r = + \widehat i + 3\widehat j + 5\widehat k (note that r→.p→=0\overrightarrow r .\overrightarrow p = 0 ) is parallel to :
A (+i^+3j^−2k^)\left( { + \widehat i + 3\widehat j - 2\widehat k} \right)
B (+i^−3j^−2k^)\left( { + \widehat i - 3\widehat j - 2\widehat k} \right)
C (−i^+3j^−2k^)\left( { - \widehat i + 3\widehat j - 2\widehat k} \right)
D (−i^−3j^+2k^)\left( { - \widehat i - 3\widehat j + 2\widehat k} \right)
Correct Answer
Option A
Solution

Since

r→\overrightarrow r

and

p→\overrightarrow p

are perpendicular to each other therefore point lies on the equitorial plane.

Therefore electric field at the point will be antiparallel to the dipole moment.

∴\therefore

E→∣∣−p→\overrightarrow E || - \overrightarrow p

⇒\Rightarrow

E→∣∣(i^+3j^−2k^)\overrightarrow E ||\left( {\widehat i + 3\widehat j - 2\widehat k} \right)
Q53
Consider two charged metallic spheres S1 and S2 of radii R1 and R2, respectively. The electric fields E1 (on S1) and E2 (on S2) on their surfaces are such that E1/E2 = R1/R2. Then the ratio V1 (on S1) / V2 (on S2) of the electrostatic potentials on each sphere is :
A (R1/R2)2
B (R2/R1)
C (R1/R2)3
D R1/R2
Correct Answer
Option A
Solution

We know, E1 =

KQ1R12{{K{Q_1}} \over {R_1^2}}

and E2 =

KQ2R22{{K{Q_2}} \over {R_2^2}}

Given

E1E2=R1R2{{{E_1}} \over {{E_2}}} = {{{R_1}} \over {{R_2}}}

⇒\Rightarrow

KQ1R12KQ2R22=R1R2{{{{K{Q_1}} \over {R_1^2}}} \over {{{K{Q_2}} \over {R_2^2}}}} = {{{R_1}} \over {{R_2}}}

⇒\Rightarrow

Q1Q2=R13R23{{{Q_1}} \over {{Q_2}}} = {{R_1^3} \over {R_2^3}}

Now

V1V2=KQ1R1KQ2R2{{{V_1}} \over {{V_2}}} = {{{{K{Q_1}} \over {{R_1}}}} \over {{{K{Q_2}} \over {{R_2}}}}}

=

R12R22{{R_1^2} \over {R_2^2}}
Q54
Given below are two statements : Statement I : A point charge is brought in an electric field. The value of electric field at a point near to the charge may increase if the charge is positive. Statement II : An electric dipole is placed in a non-uniform electric field. The net electric force on the dipole will not be zero. Choose the correct answer from the options given below :
A Both Statement I and Statement II are true.
B Both Statement I and Statement II are false.
C Statement I is true but Statement II is false.
D Statement I is false but Statement II is true.
Correct Answer
Option A
Solution

As one moves closer to a positive charge (isolated) the density of electric field line increases and so does the electric field intensity ⇒\Rightarrow Statement I is true As opposite poles of an electric dipole would experience equal and opposite forces so net force on a dipole in a uniform electric field will be zero ⇒\Rightarrow Statement II is true

Q55
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=2kλrE = {{2k\lambda } \over r}

Electric field with surface charge density

E=σ∈0E = {\sigma \over {{ \in _0}}}
2kλr=σ∈0{{2k\lambda } \over r} = {\sigma \over {{ \in _0}}}
⇒214π∈0λr=σ∈0\Rightarrow 2{1 \over {4\pi { \in _0}}}{\lambda \over r} = {\sigma \over {{ \in _0}}}
⇒82×3.14×3=σ\Rightarrow {8 \over {2 \times 3.14 \times 3}} = \sigma
⇒σ=0.424\Rightarrow \sigma = 0.424

n Cm−-2

Q56
An oil drop of radius 2 mm with a density 3g cm−-3 is held stationary under a constant electric field 3.55 ×\times 105 V m−-1 in the Millikan's oil drop experiment. What is the number of excess electrons that the oil drop will possess? (consider g = 9.81 m/s2)
A 48.8 ×\times 1011
B 1.73 ×\times 1010
C 17.3 ×\times 1010
D 1.73 ×\times 1012
Correct Answer
Option B
Solution

Fe = qE = (ne)E Fe = mg (ne)E = mg

n=mgeE=ρ43πR3×geEn = {{mg} \over {eE}} = {{\rho {4 \over 3}\pi {R^3} \times g} \over {eE}}
n=3000×43×3.14×8×10−9×9.81.6×10−19×3.55×105n = {{3000 \times {4 \over 3} \times 3.14 \times 8 \times {{10}^{ - 9}} \times 9.8} \over {1.6 \times {{10}^{ - 19}} \times 3.55 \times {{10}^5}}}
n=984704×1055.68=1.73×1010n = {{984704 \times {{10}^5}} \over {5.68}} = 1.73 \times {10^{10}}
n=1.73×1010n = 1.73 \times {10^{10}}
Q57
A certain charge Q is divided into two parts q and (Q −- q). How should the charges Q and q be divided so that q and (Q −- q) placed at a certain distance apart experience maximum electrostatic repulsion?
A Q = 2q
B Q = 4q
C Q = 3q
D Q = q2{q \over 2}
Correct Answer
Option A
Solution

Let's say the charge q and (Q −- q) are at r distance from each other.

This can be shown as According to Coulomb's law, force between both the parts can be given as

F=kq(Q−q)r2F = {{kq(Q - q)} \over {{r^2}}}
F=kr2(qQ−q2)F = {k \over {{r^2}}}(qQ - {q^2})

As we know that

dFdq=0{{dF} \over {dq}} = 0

, for maximum force.

⇒dFdq=ddq[kr2(qQ−q2)]=0\Rightarrow {{dF} \over {dq}} = {d \over {dq}}\left[ {{k \over {{r^2}}}(qQ - {q^2})} \right] = 0
⇒kr2(Q−2q)=0⇒Q=2q\Rightarrow {k \over {{r^2}}}(Q - 2q) = 0 \Rightarrow Q = 2q
Q58
An electric dipole is placed on x-axis in proximity to a line charge of linear charge density 3.0 ×\times 10−-6 C/m. Line charge is placed on z-axis and positive and negative charge of dipole is at a distance of 10 mm and 12 mm from the origin respectively. If total force of 4N is exerted on the dipole, find out the amount of positive or negative charge of the dipole.
A 0.485 mC
B 815.1 nC
C 8.8 μ\muC
D 4.44 μ\muC
Correct Answer
Option D
Solution
∣F∣=q(E1−E2)\left| F \right| = q({E_1} - {E_2})
=(q)2kλ[210×12×10−3]= (q)2k\lambda \left[ {{2 \over {10 \times 12 \times {{10}^{ - 3}}}}} \right]

⇒\Rightarrow

4=(q)×2×9×109×(3×10−6)[2120×10−3]4 = (q) \times 2 \times 9 \times {10^9} \times (3 \times {10^{ - 6}})\left[ {{2 \over {120 \times {{10}^{ - 3}}}}} \right]
⇒q=4.44\Rightarrow q = 4.44

μ\muC

Q59
Two identical tennis balls each having mass 'm' and charge 'q' are suspended from a fixed point by threads of length 'l'. What is the equilibrium separation when each thread makes a small angle 'θ\theta' with the vertical?
A x=(q2l2πε0mg)12x = {\left( {{{{q^2}l} \over {2\pi {\varepsilon _0}mg}}} \right)^{{1 \over 2}}}
B x=(q2l2πε0mg)13x = {\left( {{{{q^2}l} \over {2\pi {\varepsilon _0}mg}}} \right)^{{1 \over 3}}}
C x=(q2l22πε0m2g)13x = {\left( {{{{q^2}{l^2}} \over {2\pi {\varepsilon _0}{m^2}g}}} \right)^{{1 \over 3}}}
D x=(q2l22πε0m2g2)13x = {\left( {{{{q^2}{l^2}} \over {2\pi {\varepsilon _0}{m^2}{g^2}}}} \right)^{{1 \over 3}}}
Correct Answer
Option B
Solution
Tcos⁡θ=mgT\cos \theta = mg
Tsin⁡θ=kq2x2T\sin \theta = {{k{q^2}} \over {{x^2}}}
tan⁡θ=kq2x2mg\tan \theta = {{k{q^2}} \over {{x^2}mg}}
tan⁡θ≈sin⁡θ≈x2L\tan \theta \approx \sin \theta \approx {x \over {2L}}
x2L=Kq2x2mg{x \over {2L}} = {{K{q^2}} \over {{x^2}mg}}
x=(q2L2πε0mg)1/3x = {\left( {{{{q^2}L} \over {2\pi {\varepsilon _0}mg}}} \right)^{1/3}}
Q60
The two thin coaxial rings, each of radius 'a' and having charges +Q and −-Q respectively are separated by a distance of 's'. The potential difference between the centres of the two rings is :
A Q2πε0[1a+1s2+a2]{Q \over {2\pi {\varepsilon _0}}}\left[ {{1 \over a} + {1 \over {\sqrt {{s^2} + {a^2}} }}} \right]
B Q4πε0[1a+1s2+a2]{Q \over {4\pi {\varepsilon _0}}}\left[ {{1 \over a} + {1 \over {\sqrt {{s^2} + {a^2}} }}} \right]
C Q4πε0[1a−1s2+a2]{Q \over {4\pi {\varepsilon _0}}}\left[ {{1 \over a} - {1 \over {\sqrt {{s^2} + {a^2}} }}} \right]
D Q2πε0[1a−1s2+a2]{Q \over {2\pi {\varepsilon _0}}}\left[ {{1 \over a} - {1 \over {\sqrt {{s^2} + {a^2}} }}} \right]
Correct Answer
Option D
Solution
VA=KQa−KQa2+s2{V_A} = {{KQ} \over a} - {{KQ} \over {\sqrt {{a^2} + {s^2}} }}
VB=−KQa+KQa2+s2{V_B} = {{ - KQ} \over a} + {{KQ} \over {\sqrt {{a^2} + {s^2}} }}
VA−VB=2KQa−2KQa2+s2{V_A} - {V_B} = {{2KQ} \over a} - {{2KQ} \over {\sqrt {{a^2} + {s^2}} }}
=Q2πε0(1a−1s2+a2)= {Q \over {2\pi {\varepsilon _0}}}\left( {{1 \over a} - {1 \over {\sqrt {{s^2} + {a^2}} }}} \right)
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