question_answer

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

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

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

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

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

Correct Answer: B

Solution :

(b) \[{{V}_{A}}={{V}_{C}}\ne {{V}_{B}}\] \[{{V}_{A}}=\frac{1}{4\pi {{\varepsilon }_{0}}}\left\{ \frac{{{q}_{A}}}{a}-\frac{{{q}_{B}}}{b}+\frac{{{q}_{C}}}{c} \right\}\] \[=\frac{4\pi }{4\pi {{\varepsilon }_{0}}}\left\{ \frac{{{a}^{2}}\sigma }{a}-\frac{{{b}^{2}}\sigma }{b}+\frac{{{c}^{2}}\sigma }{c} \right\}\] \[{{V}_{A}}=\frac{1}{{{\varepsilon }_{0}}}\left\{ \frac{{{a}^{2}}\sigma }{a}-\frac{{{b}^{2}}\sigma }{b}+\frac{{{c}^{2}}\sigma }{c} \right\}\] \[{{V}_{B}}=\frac{1}{{{\varepsilon }_{0}}}\left\{ \frac{{{a}^{2}}\sigma }{a}-\frac{{{b}^{2}}\sigma }{b}+\frac{{{c}^{2}}\sigma }{c} \right\}\] \[{{V}_{C}}=\frac{1}{{{\varepsilon }_{0}}}\left\{ \frac{{{a}^{2}}\sigma }{a}-\frac{{{b}^{2}}\sigma }{b}+\frac{{{c}^{2}}\sigma }{c} \right\}\] Given, c = a + b If \[a=a,\,b=2a\]and \[c=3a\]for example, as \[c>b>a\], \[{{V}_{A}}=\frac{1}{{{\varepsilon }_{0}}}\left\{ \frac{{{a}^{2}}\sigma }{a}-\frac{4{{a}^{2}}\sigma }{2a}+\frac{{{c}^{2}}\sigma }{c} \right\}\] \[{{V}_{B}}=\frac{1}{{{\varepsilon }_{0}}}\left\{ \frac{{{a}^{2}}\sigma }{2a}-\frac{4{{a}^{2}}\sigma }{2a}+\frac{{{c}^{2}}\sigma }{c} \right\}\] \[{{V}_{C}}=\frac{1}{{{\varepsilon }_{0}}}\left\{ \frac{{{a}^{2}}\sigma }{3a}-\frac{4{{a}^{2}}\sigma }{3a}+\frac{{{c}^{2}}\sigma }{c} \right\}\] It can seen by taking out common factors that \[{{V}_{A}}={{V}_{C}}>{{V}_{B}}\] i.e., \[{{V}_{A}}={{V}_{C}}\ne {{V}_{B}}\]