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A charge Q is distributed over two concentric hollow, spheres of radius r and R(

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a

QR+r

b

Q(R+r)4πe0R2+r2

c

QR2+r24πε0(R+r)

d

Zero

answer is B.

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Detailed Solution

since, the surface densities are equal, henceq14πr2=q24πR2(where q1+q2=Q)or q1r2=q2R2=q1+q2r2+R2=Qr2+R2∴q1=Qr2+R2×r2 and q2=Qr2+R2×R2So, potential at the common centre,V=q14πε0r+q24πε0R=14πε0q1r+q2R=14πε0QR2+r2×r2r+QR2+r2×R2R=14πε0Q(R+r)R2+r2

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