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Two identical thin ring, each of radius ...

Two identical thin ring, each of radius R meters, are coaxially placed a distance R metres apart. If `Q_1` coulomb, and `Q_2` coulomb, are repectively the charges uniformly spread on the two rings, the work done in moving a charge q from the centre of one ring to that of the other is

A

zero

B

`(q(Q_(1) - Q_(2))(sqrt(2) -1))/(sqrt(2)( 4 pi epsi_(0)R))`

C

`(qsqrt(2)(Q_(1) + Q_(2)))/((4 pi epsi_(0)R))`

D

`q(Q_(1)//Q_(2))(sqrt(2)+1)(4pi epsi_(0) R)`

Text Solution

Verified by Experts

The correct Answer is:
B
`V_(C_1)=V_(Q_1)+V_(Q_2)=(1)/(4piepsi_(0))(Q_1)/(R)+(1)/(4piepsi_(0))(Q_2)/(Rsqrt(2))=(1)/(4piepsi_(0)R)(Q_(1)+(Q_2)/(sqrt2))`
Similarly `V_(C_2)=(1)/(4piepsi_(0)R)(Q_(2)+(Q_1)/(sqrt2))" "therefore" "DeltaV=V_(C_1)-V_(C_2)`
`=(1)/(4piepsi_(0)R)[Q_(1)-Q_(2)-(1)/(sqrt2)(Q_(1)-Q_(2))]=(Q_(1)-Q_(2))/(sqrt(2)(4piepsi_(0)R))(sqrt(2)-1)`
`W=qDeltaV=q(Q_(1)-Q_(2))(sqrt(2)-1)//sqrt(2)(4piepsi_(0)R)`
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