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A charge +q is fixed at each of the poin...

A charge `+q` is fixed at each of the points `x = x_0, x = 3x_0, x = 5x_0,` and so on, on the x-axis, and a charge `-q` is fixed at each of the points ` x= 2x_0, x = 4x_0, x = 6x_0,` and so on. Here, `x_0` is a positive constant. Take the electric potential at a point due to 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 charges is

A

zero

B

`q/(8piepsilon_(0)x_(0)ln2)`

C

infinite

D

`(qln(2))/(4piepsilon_(0)x_(0))`

Text Solution

Verified by Experts

The correct Answer is:
D
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)`
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