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Question

A charge q is distributed over two concentric hollow conducting sphere of radii r and R(>r) such that their surface charge densities are equal. The potential at their common centre is

A
Zero
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B
q4πϵ0(r+R)(r2+R2)2
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C
q4πϵ0(1r+1R)
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D
Vc=q4πϵ0(r+Rr2+R2)
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Solution

The correct option is D Vc=q4πϵ0(r+Rr2+R2)
Let charge on inner sphere be q1 and, on outer shell be q2.

According to the given data in the problem,
q=q1+q2 (1)

Since the surface charge density (σ) on both the spheres are equal, so

σ=q14πr2=q24πR2

q1q2=r2R2 ........(2)

Adding +1 on both sides we get,

q1+q2q2=r2+R2R2

From equation (1),

qq2=r2+R2R2

q2=qR2r2+R2...(3)

Substituting this in (2),

q1=qr2r2+R2...(4)

The potential at the common centre is

Vc=V1+V2

Vc=14πϵ0[q1r+q2R]

Substituting the values from (3) and (4)

Vc=14πϵ0[(qr2+R2)[r2r+R2R]]

Vc=q4πϵ0(r+Rr2+R2)

Hence, option (d) is the correct answer.

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