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Two conducting spheres of radius R and 3...

Two conducting spheres of radius R and 3R carry charges Q and `-2 Q`. Between these spheres a neutral conducting sphere of radius 2R is connected. The separation between the sphere is considerably large Charge flows through conducting wire due to potential difference.

The final electric potential of sphere of radius 3R will be :

A

`-(kQ)/(6 R)`

B

`-(kQ)/(2R)`

C

`-(2kQ)/(3R)`

D

`-(3kQ)/(R)`

Text Solution

Verified by Experts

The correct Answer is:
A

Final potential of spheres will be same
So, `K(x)/(R)=(Ky)/(2R)=(K(-Q-x-y))/(3R)`
`y=2x`
and `3x =-Q-x-y`
`:. 6x =-Q`
`x = (Q)/(6) rArr y = -(Q)/(3)`
Charge on sphere of radius 3R is `- (Q)/(2)`
Change in potential energy of sphere of radius 'R' is
`DeltaU =(KQ^(2))/(2r)-(K(-Q//6)^(2))/(2r)rArr Delta U=(35 KQ^(2))/(72 R)`.
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