Three concentric spherical shells have radii a, b, and c(a < b < c) and have surface charge densities σ, −σ andσ respectively. If VA, VB and VC denote the potentials of the three shells, then, for c = a + b, we have :
A
VC = VB = VA
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B
VC = VA≠VB
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C
VC = VB≠VA
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D
VC≠VB≠VA
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Solution
The correct option is BVC = VA≠VB It is clear from figure that Va=σaε0−σbε0+σcε0=σε0(a−b+c)c=a+b As c=a+b ∴VA=σε0(2a) VB=σε0[a2b−b+c]=σε0[a2b−b+a+b] VB=σε0[a2b+a]=σaε0(a+b)b=σ0acε0b VC=σε0[a2b−a2c+c]=σaε0[a2−b2+c2c]=σε0[(a+b)(a−b)+(a+b)2c] VC=σε0[(a+b)c(a−b+a+b)]=σ2aε0 we found that, VC = VA≠VB