Two thin wire rings each having radius R are placed at distance d apart with their axes coinciding. The charges on the two are `+Q` and `-Q`. The potential difference between the centre so the two rings is

Options

A. `q^R/(4pi epsilon_0d^2)`
B. `q/(2pi epsilon_0)[1/R-1/sqrt(R^2+d^2)]`
C. Zero
D. `q/(\4pi epsilon_0)[1/R-1/sqrt(R^2+d^2)]`

Answer

B

Explanation

`V_A`= Potential due to charge +q on ring A+ Potential due to change -q on ring B
`=1/(4 pi e_0) (q/R- q/sqrt(R^2+d^2))`
Similarly `V_B=1/(4 pi e_0) (-q/R +q/sqrt(R^2+a^2))` Potential difference `V_A-V_B`
`1/(4 pi e_0) (q/R- q/sqrt(R^2+q^2)) -1/(4 pi e_0) q/sqrt(R^2+d^2) =q/(2 pi e_0) (1/R-1/sqrt(R^2+d^2))`