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.

Options

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)`

Answer

B

Explanation

`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)`