JEE Main · 2013 · 07 Apr 2013 · Offline · Electrostatics

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A charge Q is uniformly distributed over a long rod AB of length L as shown in the figure. The electric potential at the point O lying at a distance L from the end A is:

(A) $\frac{3 Q}{4 π ε _{0} L}$
(B) $\frac{Q}{4 π ε _{0} L l n 2}$
(C) $\frac{Q l n 2}{4 π ε _{0} L}$ ✓ Correct
(D) $\frac{Q}{8 π ε _{0} L}$

Answer: (C)

Solution

Spread the charge uniformly as

λ = Q / L

and take an element

d x

at distance

x

from

O

, where

x

runs from

L

2 L

since

O

lies a distance

L

from the near end

A

. Each element contributes

d V = \frac{k λ d x}{x}

, and integrating,

V = \frac{k Q}{L} \int_{L}^{2 L} \frac{d x}{x} = \frac{k Q}{L} ln 2 = \frac{Q ln 2}{4 π ε _{0} L}

JEE Main · 2013 · 07 Apr 2013 · Offline · Single correct option

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