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Q.

A cylindrical capacitor of length ‘l’ consists of a solid conducting cylinder of radius ‘a’ which carries a linear charge density $+ λ$ , surrounded by a concentric outer cylindrical shell of radius ‘b’ which carries a linear charge density $- λ$ . The capacitance of the capacitor is

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a

$\frac{4 π \in_{0} a}{\ln (\frac{b}{a})}$

b

$\frac{2 π \in_{0} b}{\ln (\frac{b}{a})}$

c

$\frac{4 π \in_{0} l}{\ln (\frac{b}{a})}$

d

$\frac{2 π \in_{0} l}{\ln (\frac{b}{a})}$

answer is A.

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Detailed Solution

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The electric field at a distance r from a conducting cylinder is given by
$E = \frac{λ}{2 π ϵ_{0} r}$
Therefore, the potential difference between the inner cylinder and the outer cylinder is
$V = - \int E d r = - \frac{λ}{2 π \in_{0}} \int_{b}^{a} \frac{d r}{r}$

$= \frac{λ}{2 π \in_{0}} \int_{a}^{b} \frac{d r}{r} \Rightarrow V = \frac{λ}{2 π \in_{0}} \ln (\frac{b}{a})$

$Now C = \frac{Q}{V} = \frac{λ l}{\frac{λ}{2 π \in_{0}} \ln (\frac{b}{a})}$

$C = \frac{2 π \in_{0} l}{\ln (\frac{b}{a})}$

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