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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_(0)R)`

C

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

D

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

Text Solution

Verified by Experts

The correct Answer is:
B

`V_(1)=(Q_(1))/(4pi epsilon_(0)R)+(Q_(2))/(4pi epsilon_(0) sqrt(2)R)`
and `V_(2)=(Q_(1))/(4pi epsilon_(0) sqrt(2)R)+(Q_(2))/(4pi epsilon_(0)R)`
`W_(1 to 2)=q(V_(2)-V_(1))`
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