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Two identical thin rings, each of radius...

Two identical thin rings, each of radius R, are coaxially placed at a distance R. If `Q_(1) and Q_(2)` are respectively, the charges uniformly spread on the two rings, find the work done in moving a charge q from centre of ring having charge `Q_(1)` to the other ring.

A

zero

B

`(q(Q_1-Q_2)(sqrt2-1))/(4sqrt(2)piepsilon_0R)`

C

`(qsqrt(2)(Q_1+Q_2))/(4piepsilon_0R)`

D

`(q(Q_1+Q_2)(sqrt2+1))/(4sqrt(2)piepsilon_0R)`

Text Solution

Verified by Experts


the work done in moving a charge from A to B is
`W=(TPE)_(A)-(TPE)_(B)`
`TPE=` total potential energy
`(TPE)_(A)=[PE" due to "Q_(1)+PE" due to "Q_(2)]`
`=[((Q_(1))/(4piepsilon_(0)R))xxq+((Q)/(4piepsilon_(0)sqrt(R^(2)+R^(2))))q]`
`=(q)/(4piepsilon_(0)R)[Q_(1)+(Q_(2))/(sqrt(2))]`
`(TPE)_(B)[PE" due to "Q_(2)+PE" due to "Q_(2)]`
`=[((Q_(2))/(4piepsilon_(0)R))q+((Q_(1))/(4piepsilon_(0)sqrt(R^(2)+R^(2))))q]`
`=(q)/(4piepsilon_(0)R)[Q_(2)+(Q_(1))/(sqrt(2))]`
`thereforeW=(q)/(4piepsilon_(0)R)[Q_(1)(1-(1)/(sqrt(2)))-Q_(2)(1-(1)/(sqrt(2)))]`
`=(q(Q_(1)-Q_(2)))/(4piepsilon_(0)R)((sqrt(2)-1)/(sqrt(2)))`
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