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. `oo`
C. `(q)/(4pi in_(0) x_(0) in 2)`
D. `(q IN 2)/(4pi in_(0) x_(0) in 2)`

Answer

D

Explanation

Potential due to positive charges
`V_(1) = (1)/(4pi in_(0)) (q)/(x_(0)) + (1)/(4pi in_(0)) (q)/(3x_(0)) + …..`
`= (q)/(4pi in_(0) x_(0)) [1 + (1)/(3) + …..]`
Potential due to negative charge
`V_(2) = (1)/(4pi in_(0)) (-q)/(2x_(0)) + (1)/(4pi in_(0)) (-q)/(4x_(0)) + ....`
`= (+q)/(4pi in_(0) x_(0)) [ - (1)/(2) - (1)/(4) - (1)/(6) + .....]`
Totol potential `V = V_(1) + V_(2)`
`(q)/(4pi in_(0) x_(0)) [1 - (1)/(2) + (1)/(3) - (1)/(4) + ......]`
`[1 - (1)/(2) + (1)/(3) - (1)/(4) + ........ = In2]`
`V = (q In2)/(4pi in_(0) x_(0))`