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Question

Physics Question on electrostatic potential and capacitance

A charge + q is fixed at each of the points x=x0,x=3x0,x=5x0...8x = x_0,x=3x_0,x=5x_0...8 on the x-axis and a charge - q is fixed at each of the points x=2x0,x=4x0,x=6x0...,8x = 2x_0,x=4x_0,x=6x_0...,8 Here, x0x_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/4pε0r.Q/4 p\varepsilon_0r. Then the potential at the origin due to the above system of charges is

A

zero

B

q8pε0x0In2\frac{q}{8 p\varepsilon_0x_0\, In\, 2}

C

infinite

D

qIn24pε0x0\frac{q\, In\, 2}{4 p\varepsilon_0x_0}

Answer

qIn24pε0x0\frac{q\, In\, 2}{4 p\varepsilon_0x_0}

Explanation

Solution

Potential at origin will be given by
V=q4pε0[1x012x0+13x014x0+....]V=\frac{q}{4 p\varepsilon_0}\bigg[\frac{1}{x_0}-\frac{1}{2x_0}+\frac{1}{3x_0}-\frac{1}{4x_0}+....\bigg]
=q4pε0.1x0[112+1314+....]=q4pε0x0In(2)\, \, \, \, =\frac{q}{4 p\varepsilon_0}.\frac{1}{x_0}\bigg[1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+....\bigg]=\frac{q}{4 p\varepsilon_0 x_0}In\, (2)