Question
Three concentric spherical shells have radii $$a,b$$ and $$c\left( {a < b < c} \right)$$ 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
A.
$${V_C} = {V_A} \ne {V_B}$$
B.
$${V_C} = {V_B} \ne {V_A}$$
C.
$${V_C} \ne {V_B} \ne {V_A}$$
D.
$${V_C} = {V_B} = {V_A}$$
Answer :
$${V_C} = {V_B} = {V_A}$$
Solution :
$$\eqalign{
& {V_A} = \frac{1}{{4\pi {\varepsilon _0}}}\frac{{\sigma 4\pi {a^2}}}{a} - \frac{1}{{4\pi {\varepsilon _0}}}\frac{{\sigma 4\pi {b^2}}}{b} + \frac{1}{{4\pi {\varepsilon _0}}} \cdot \frac{{\sigma 4\pi {c^2}}}{c} \cr
& = \frac{\sigma }{{{\varepsilon _0}}}\left( {a - b + c} \right) = \frac{\sigma }{{{\varepsilon _0}}}\left( {2a} \right)\,\,\,\,\left( {\because c = a + b} \right) \cr
& {V_B} = \frac{1}{{4\pi {\varepsilon _0}}} \cdot \frac{{\sigma 4\pi {a^2}}}{c} - \frac{1}{{4\pi {\varepsilon _0}}}\frac{{\sigma 4\pi {b^2}}}{b} + \frac{1}{{4\pi {\varepsilon _0}}} \cdot \frac{{\sigma 4\pi {c^2}}}{c} \cr
& = \frac{\sigma }{{{\varepsilon _0}}}\left( {\frac{{{a^2}}}{c} - b + c} \right) = \frac{\sigma }{{{\varepsilon _0}}}\left( {2a} \right)\,\,\,\,\left( {\because c = a + b} \right) \cr} $$
$$\eqalign{
& {\text{and}}\,\,{V_C} = \frac{1}{{4\pi {\varepsilon _0}}} \cdot \frac{{\sigma 4\pi {a^2}}}{c} - \frac{1}{{4\pi {\varepsilon _0}}}\frac{{\sigma 4\pi {b^2}}}{c} + \frac{1}{{4\pi {\varepsilon _0}}} \cdot \frac{{\sigma 4\pi {c^2}}}{c} \cr
& = \frac{\sigma }{{{\varepsilon _0}}}\left( {\frac{{{a^2}}}{c} - \frac{{{b^2}}}{c} + c} \right) = \frac{\sigma }{{{\varepsilon _0}}}\left( {2a} \right)\,\,\,\,\left( {\because c = a + b} \right) \cr
& {\text{Hence,}}\,\,{V_A} = {V_C} = {V_B} \cr} $$