Electrostatics

JEE Main level, single correct, medium

209 of 311

Question 209 of 311

Three concentric metal shells A,BA,B and CC of respective radii a,ba,b and c(a<b<c)c(a\lt b\lt c) have surface charge densities <c) charge="" densities="" have="" math="" surface=""> +σ,−σ+\sigma ,-\sigma and +σ+\sigma respectively. The potential of shell BB is:</c)>

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Answer: (C) σϵ0[a2−b2b+c]\dfrac{\sigma }{{\epsilon }_{0}}\left[\dfrac{{a}^{2}-{b}^{2}}{b}+c\right]



Net potential at B is due to the superposition of potential due to all shells,


∨B=∨B(I)+∨B(II)+∨B(III){∨}_{B}={∨}_{B\left(I\right)}+{∨}_{B\left(II\right)}+{∨}_{B\left(III\right)}

=K4πa2σb+K4πb2(−σ)b+K4πc2(σ)c=\dfrac{K4\pi {a}^{2}\sigma }{b}+\dfrac{K4\pi {b}^{2}\left(-\sigma \right)}{b}+\dfrac{K4\pi {c}^{2}\left(\sigma \right)}{c}

Required potential of b


=σa2ϵ0b−σb2ϵ0b+σc2ϵ0c=\dfrac{\sigma {a}^{2}}{{\epsilon }_{0}b}-\dfrac{\sigma {b}^{2}}{{\epsilon }_{0}b}+\dfrac{\sigma {c}^{2}}{{\epsilon }_{0}c}

=σϵ0(a2−b2b+c)=\dfrac{\sigma }{{\epsilon }_{0}}\left(\dfrac{{{a}^{2}-b}^{2}}{b}+c\right)

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