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)/(8 pi epsilon_(0)x_(0)ln2)`
C. `oo`
D. `(q ln 2)/(4 pi epsilon_(0)x_(0))`

Answer

D

Explanation

`V = (1)/(4pi epsilon_(0))[(q)/(x_(0)) - (q)/(2x_(0)) + (q)/(3x_(0)) - (q)/(4x_(0))…..]`
`= (1)/(4pi epsilon_(0))(q)/(x_(0)) [1-(1)/(2) + (1)/(3) - (1)/(4)] = (1)/(4pi epsilon_(0))(q)/(x_(0)) ln2`