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Question

Two identical thin rings each of radius R meters are coaxially placed at a distance R meters apart. If Q1 coulomb and Q2 coulomb are respectively 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 other is

A
Zero
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B
q(Q2Q1)(21)2.4πϵ0R
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C
q2(Q1+Q2)4πϵ0R
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D
q(Q1+Q2)(2+1)2.4πϵ0R
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Solution

The correct option is B q(Q2Q1)(21)2.4πϵ0R

W=q(Vσ2Vσ1)
where Vσ1=Q14πϵR+Q24πϵ0R2
and Vσ2=Q24πϵR+Q14πϵ0R2
Vσ2Vσ1=(Q2Q1)4πϵ0R[112]
So, W=q.(Q2Q1)4πϵ0R(21)2

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