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. zero
B. `q/(8piepsilon_(0)x_(0)ln2)`
C. infinite
D. `(qln(2))/(4piepsilon_(0)x_(0))`

Answer

D

Explanation

Potental at origin will be given by
`V=q/(4piepsilon_(0))[1/x_(0)-1/(2x_(0))+1/(3x_(0))-1/4x_(0)+…]`
`=q/(4pi epsilon_(0))(1/x_(0))[1-1/2+1/3-1/4+…]`
`=q/(4piepsilon_(0)x_(0))ln(2)`