A charge `+q` is fixed at each of the points `x=x_0`, `x=3x_0`, `x=5x_0`,…………`x=oo` on the x axis, and a charge `-q` is fixed at each of the points `x=2x_0`, `x=4x_0`, `x=6x_0`, …………`x=oo`. 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//(4piepsilon_0r)`.Then, the potential at the origin due to the above system of

Options

A. `0`
B. `(q 1n2)/(8 pi epsilon_(0) x_(0))`
C. `oo`
D. `(q 1n 2)/(4 pi epsilon_(0) x_(0))`

Answer

4

Explanation

`V_(0) = (1)/(4 pi in_(0)) [ (q)/(x_(0)) + (-q)/(2x_(0)) + (q)/(3x_(0)) + …oo]`
`= (1)/(4 pi in_(0)) .(q)/(x_(0)) [ 1 - (1)/(2) + (1)/(3) - …oo]`
`= (1)/(4 pi in_(0)) .(q)/(x_(0)) 1n 2 = (q 1n 2)/(4 pi in_(0) x_(0))`