Three concentric spherical shells have radii a, b and c (a < b < c) and have surface charge densities σ,−σ and σ respectively. If VA, VB and VC denotes the potentials of the three shells, then for c = a + b, we have [NEET 2009]

Three concentric spherical shells have radii a, b and c (a < b < c) and have surface charge densities σ,−σ and σ respectively. If VA, VB and VC denotes the potentials of the three shells, then for c = a + b, we have [NEET 2009]

a) $V_C=V_B\neq V_A$

b) $V_C\neq V_B\neq V_A$

c) $V_C=V_B=V_A$

d) $V_C=V_A\neq V_B$

Concept: Use Van de graph principles. In concentric spheres, the voltage due to the charges on the surface and the voltage due to the inner spheres and outer spheres should be added. $V_{inside\hspace{2mm}sphere}= V_{at\hspace{2mm}surface\hspace{2mm}of\hspace{2mm}sphere}$

\[V_A=\frac{1}{4\pi\epsilon_o}\frac{q_A}{r_A}-\frac{1}{4\pi\epsilon_o}\frac{q_B}{r_B}+\frac{1}{4\pi\epsilon_o}\frac{q_C}{r_C}\]

\[=K\frac{\sigma 4\pi a^2}{a}-K\frac{\sigma 4\pi b^2}{b}+K\frac{\sigma 4\pi c^2}{c}\]

\[=K\sigma 4\pi \hspace{2mm}(a-b+c)\]

\[=K\sigma 4\pi \hspace{2mm}(a-b+a+b)\]

\[=K \sigma 4\pi \hspace{2mm} (2a) \]

\[V_B=\frac{1}{4\pi\epsilon_o}\frac{q_A}{r_B}-\frac{1}{4\pi\epsilon_o}\frac{q_B}{r_B}+\frac{1}{4\pi\epsilon_o}\frac{q_C}{r_C}\]

\[=K\frac{\sigma 4\pi a^2}{b}-K\frac{\sigma 4\pi b^2}{b}+K\frac{\sigma 4\pi c^2}{c}\]

\[=K\sigma 4\pi \hspace{2mm}(\frac{a^2}{b}-b+c)\]

\[=K\sigma 4\pi \hspace{2mm}(\frac{a^2}{b}-b+a+b)\]

\[=K \sigma 4\pi \hspace{2mm} (\frac{a^2 +ab}{b}) \]

\[V_C=\frac{1}{4\pi\epsilon_o}\frac{q_A}{r_C}-\frac{1}{4\pi\epsilon_o}\frac{q_B}{r_C}+\frac{1}{4\pi\epsilon_o}\frac{q_C}{r_C}\]

\[=K\frac{\sigma 4\pi a^2}{c}-K\frac{\sigma 4\pi b^2}{c}+K\frac{\sigma 4\pi c^2}{c}\]

\[=K\sigma 4\pi \hspace{2mm}(\frac{a^2}{c}-\frac{b^2}{c}+c)\]

\[=K \sigma 4\pi \hspace{2mm} (2a) \]

So the answer is \[V_C=V_A\neq V_B\]