question_answer

Three concentric spherical shells have radii a, b and \[c(a<b<c)\] and have surface charge densities   \[\sigma ,\,-\sigma \] and \[\sigma \] respectively.   If \[{{V}_{A}},\,{{V}_{B}}\] and \[{{V}_{C}}\] denote the potentials of the three shells, then for \[C=a+b,\] we have [AIPMT (S) 2009]

A) \[{{V}_{A}}={{V}_{C}}\ne {{V}_{B}}\]

B) \[{{V}_{C}}={{V}_{B}}\ne {{V}_{A}}\]

C) \[{{V}_{C}}\ne {{V}_{B}}\ne {{V}_{A}}\]

D) \[{{V}_{C}}={{V}_{B}}={{V}_{A}}\]

Correct Answer: A

Solution :

[a] Hence, \[{{V}_{A}}=\frac{1}{4\pi {{\varepsilon }_{0}}}\,\frac{\sigma 4\pi {{a}^{2}}}{a}-\frac{1}{4\pi {{\varepsilon }_{0}}}\,\frac{\sigma 4\pi {{b}^{2}}}{b}\]

\[+\frac{1}{4\,\pi {{\varepsilon }_{0}}},\,\,\frac{\sigma 4\pi {{c}^{2}}}{c}\]

\[=\frac{\sigma }{{{\varepsilon }_{0}}}(a-b+c)=\frac{\sigma }{{{\varepsilon }_{0}}}(2a)\,\]\[(\because \,c=a+b)\]

\[{{V}_{B}}=\frac{1}{4\pi {{\varepsilon }_{0}}}\,-\,\frac{\sigma 4\pi {{\alpha }^{2}}}{a}\,-\frac{1}{4\pi {{\varepsilon }_{0}}}\,\frac{\sigma 4\pi {{b}^{2}}}{b}\]

\[+\frac{1}{4\,\pi {{\varepsilon }_{0}}},\,\,\frac{\sigma 4\pi {{c}^{2}}}{c}\]

\[=\frac{\sigma }{{{\varepsilon }_{0}}}\,\left( \frac{{{a}^{2}}}{c}-b+c \right)\,=\frac{\sigma }{{{\varepsilon }_{0}}}(2a)\,\,(\because \,c=a+b)\]

And      \[{{V}_{C}}=\frac{1}{4\pi {{\varepsilon }_{0}}}\cdot \frac{\sigma \,4\,\pi {{\sigma }^{2}}}{c}-\frac{1}{4\,\pi {{\varepsilon }_{0}}}\frac{\sigma \,4\,\pi \,{{b}^{2}}}{c}\]

\[+\frac{1}{4\,\pi {{\varepsilon }_{0}}},\,\,\frac{\sigma 4\pi {{c}^{2}}}{c}\]

\[=\frac{\sigma }{{{\varepsilon }_{0}}}\,\left( \frac{{{a}^{2}}}{c}-\frac{{{b}^{2}}}{c}+c \right)\,=\frac{\sigma }{{{\varepsilon }_{0}}}(2a)\,\,(\because \,c=a+b)\]

Hence,  \[{{V}_{A}}={{V}_{C}}\ne {{V}_{B}}\]