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A charge $-q$ is fixed at each of the points $x= 2 m,x=4m,x=6m.... $ up to infinity and charge $+q$ is fixed at each of the points $x=1 m ,x= 3m , x=5m....$ up to infinity. The potential at the origin will be :

$(A)\;zero\\ (B)\;\frac{q}{4\pi \in _0 ln (2)} \\ (C)\; \frac{q ln(2)}{4\pi \in _0} \\ (D)\;Infinite $

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$V= \large\frac{1}{4 \pi \in _0} \bigg\{ \large\frac{q}{1} +\frac{q}{3} +\frac{q}{5} +.....\bigg\}+ \frac{1}{4 \pi \in _0} \bigg\{ \frac{-q}{2}-\frac{q}{4}-\frac{q}{6}+....\bigg\}$
$\qquad= \large\frac{1}{4 \pi \in _0 } \bigg\{1-\frac{1}{2} +\frac{1}{3} -\frac{1}{4} +\frac{1}{5} -\frac{1}{6} +.....\bigg\}$
=> $\large \frac{1}{4\pi \in _0}$$ q ln (1+1)$
=> $ \large \frac{q ln(2)}{4\pi \in _0}$
Hence C is the correct answer.
answered Jan 15, 2014 by meena.p
 

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