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  4. A charge + q is fixed at each of the points x = x0,x=3x0,x=5x0...∞ on the x-axis and a charge - q is fixed at each of the points x = 2x0,x=4x0,x=6x0...,∞ Here, x0 is a positive constant. Take the electric potential at a point due to a charge Q at a distance r from it to be Q/4 π ε0 r. Then the potential at the origin due to the above system of charges is

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Q. A charge + q is fixed at each of the points $x = x_0,x=3x_0,x=5x_0...\infty$ on the x-axis and a charge - q is fixed at each of the points $x = 2x_0,x=4x_0,x=6x_0...,\infty$
Here, $x_0$ is a positive constant. Take the electric potential at a point due to a charge Q at a distance r from it to be $Q/4 \pi \varepsilon_0 r.$
Then the potential at the origin due to the above system of charges is

IIT JEEIIT JEE 1998Electrostatic Potential and Capacitance

Solution:

Potential at origin will be given by
$V=\frac{q}{4 \pi \varepsilon_0} [\frac{1}{x_0}-\frac{1}{2x_0}+\frac{1}{3x_0}-\frac{1}{4x_0}+....]$
$ =\frac{q}{4 \pi \varepsilon_0}.\frac{1}{x_0}[1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+....]=\frac{q}{4 \pi \varepsilon_0 x_0}\ln\, (2)$