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Question

Three concentric spherical shells have radii a, b and c(a < b < c) and have surface charge densities +σ,σ,+σ respectively. If VA,VB and VC denote the potentials of three shells, then for c=a+b, we have

A
VC=VB=VA
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B
VC=VBVA
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C
VCVB=VA
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D
VCVBVA
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Solution

The correct option is C VCVB=VA

The concentric spherical shells have radii =a,b and c(a<b<c) and have surface charge densities =+6,6,+6 respectively.
If VA, VB, VC denote the potentials of three shells then for c=a+b we have

V= potential for a concentric shell [14πϵ0]qr(1)

σ= surface charge density =qA=q4πϵ0q4πr2(2)

from (1), VA=14πϵ0[(qAa)+qBb+qCc]

from (2), q=6 4πr2 also,

σA=σ

σB=σ

σC=σ

VA=14πϵ0[4πσa2a4πσb2b+4πσc2c](3)

VA=1ϵ0[σaσb+σc]

=σϵ0(ab+c)(3)
Similarly,

VB=[14πϵ0][qAa+qBb+qCc]

=14πϵ0[4πσa2a4πσb2b+4πσc2c]

VB=σϵ0[a2bb+c](4)

VC=σϵ0[a2cb2c+c](5)
Putting c=a+b

VA=VC>VB

i.e VA=VCVB

1245004_945415_ans_62468798cfd544139ae83644878ad68a.jpg

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