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Two thin wire rings each having radius R...

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

A

zero

B

`(QR)/(4pi varepsilon_0d^2)`

C

` Q/( 4 pi varepsilon_0) [1/R - 1/(sqrt R^2 +d^2)]`

D

`(QR)/(4pi varepsilon_0d^2)`

Text Solution

Verified by Experts

The correct Answer is:
D
`V_A =` (potnyial due to chatge `+q` on ring `A 0 +` (potential due to charge ` -q` on ring `B`)
`V_A = (v+q A)`
`= 1/(4 pi varepsilon_0) (q/R - q/d_1), d_1 = sqrt (R^2 +d^2)`
`= 1/(4 pivarepsilon_0 ) q/R - q/(sqrt (R^2 +d^2))` ...(1)

Similarly,
`V_B = 1/( 4 pi varepsilon_0) (- q/R + q/(sqrt (R^2 +d^2))`
Potential difference `V_A - V_B`
` = 1/(4 pivarepsilon_0) q/R + 1/(4 pi varepsilon_0) q/R - 1/(4 pi varepsilon_0) q/(sqrt (R^2 +d^2)) - 1/(4 pi varepsilon_0) q/(sqrt (R^3 +d^2)) - 1/(4 pi varepsilon_0) q/(sqrt R^2 +d^2))`.
`= 1/(2 pi varepsilon_0) (q)/R - q/(sqrt (R^2 +d^2)`.
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