Question

A charge $Q$ is distributed over three concentric spherical shells of radii $a,b,c\left( {a < b < c} \right)$ such that their surface charge densities are equal to one another. The total potential at a point at a distance $r$ from their common center, where $r < a$ , would be
(A) $\dfrac{Q}{{4\pi {\varepsilon _o}\left( {a + b + c} \right)}}$
(B) $\dfrac{{Q\left( {a + b + c} \right)}}{{4\pi {\varepsilon _o}\left( {{a^2} + {b^2} + {c^2}} \right)}}$
(C) $\dfrac{Q}{{12\pi {\varepsilon _o}}}\dfrac{{ab + bc + ca}}{{abc}}$
(D) $\dfrac{Q}{{4\pi {\varepsilon _o}}}\dfrac{{\left( {{a^2} + {b^2} + {c^2}} \right)}}{{\left( {{a^3} + {b^3} + {c^3}} \right)}}$

Answer 1

Hint Since in the question we are given that the surface charge densities are the same, so from the equation of the surface charge densities we need to find the charge on each conductor. Then in the formula for the potential, we substitute the value of the charge and then calculate to get the answer.

Formula Used In the solution we will be using the following formula
$\Rightarrow \sigma = \dfrac{q}{{4\pi {r^2}}}$
where $\sigma$ is the surface charge density on a sphere
$q$ is the charge and $4\pi {r^2}$ is the surface area of the sphere.
$\Rightarrow V = \dfrac{q}{{4\pi {\varepsilon _o}r}}$
where $V$ is the potential and $r$ is the radius of the sphere.

Complete step by step answer
In the question we are given that the total charge on the three spheres is ${q_a} + {q_b} + {q_c} = Q$ and the surface charge densities of the three spheres are equal. So ${\sigma _a} = {\sigma _b} = {\sigma _c} = \sigma$
Now the surface charge density of the three spheres will be given from the formula, $\sigma = \dfrac{q}{{4\pi {r^2}}}$
So we get
$\Rightarrow {\sigma _a} = \dfrac{{{q_a}}}{{4\pi {a^2}}}$ , ${\sigma _b} = \dfrac{{{q_b}}}{{4\pi {b^2}}}$ and ${\sigma _c} = \dfrac{{{q_c}}}{{4\pi {c^2}}}$
From here we get,
$\Rightarrow {q_a} = 4\pi {a^2}{\sigma _a}$ , ${q_b} = 4\pi {b^2}{\sigma _b}$ and ${q_c} = 4\pi {c^2}{\sigma _c}$
Now ${q_a} + {q_b} + {q_c} = Q$ . So substituting we get,
$\Rightarrow 4\pi {a^2}{\sigma _a} + 4\pi {b^2}{\sigma _b} + 4\pi {c^2}{\sigma _c} = Q$ .
As ${\sigma _a} = {\sigma _b} = {\sigma _c} = \sigma$ , so $4\pi {a^2}\sigma + 4\pi {b^2}\sigma + 4\pi {c^2}\sigma = Q$
Now on taking common, $4\pi \sigma \left( {{a^2} + {b^2} + {c^2}} \right) = Q$
So we have $4\pi \sigma = \dfrac{Q}{{\left( {{a^2} + {b^2} + {c^2}} \right)}}$
Now the potential at a point which is inside the sphere is given by, $V = \dfrac{q}{{4\pi {\varepsilon _o}r}}$ . So the potential at the point due to the sphere a is, ${V_a} = \dfrac{{{q_a}}}{{4\pi {\varepsilon _o}a}}$ . Similarly for sphere b and c is ${V_b} = \dfrac{{{q_b}}}{{4\pi {\varepsilon _o}b}}$ and ${V_c} = \dfrac{{{q_c}}}{{4\pi {\varepsilon _o}c}}$
So the total potential is the sum of these three potentials.
$\Rightarrow V = {V_a} + {V_b} + {V_c} = \dfrac{{{q_a}}}{{4\pi {\varepsilon _o}a}} + \dfrac{{{q_b}}}{{4\pi {\varepsilon _o}b}} + \dfrac{{{q_c}}}{{4\pi {\varepsilon _o}c}}$
Substituting ${q_a} = 4\pi {a^2}{\sigma _a}$ , ${q_b} = 4\pi {b^2}{\sigma _b}$ and ${q_c} = 4\pi {c^2}{\sigma _c}$ we get
$\Rightarrow V = \dfrac{{4\pi {a^2}{\sigma _a}}}{{4\pi {\varepsilon _o}a}} + \dfrac{{4\pi {b^2}{\sigma _b}}}{{4\pi {\varepsilon _o}b}} + \dfrac{{4\pi {c^2}{\sigma _c}}}{{4\pi {\varepsilon _o}c}}$
As ${\sigma _a} = {\sigma _b} = {\sigma _c} = \sigma$ and taking common we have,
$\Rightarrow V = \dfrac{{4\pi \sigma \left( {a + b + c} \right)}}{{4\pi {\varepsilon _o}}}$
Substituting $4\pi \sigma = \dfrac{Q}{{\left( {{a^2} + {b^2} + {c^2}} \right)}}$ in the above equation,
$\Rightarrow V = \dfrac{Q}{{4\pi {\varepsilon _o}}}\dfrac{{\left( {a + b + c} \right)}}{{\left( {{a^2} + {b^2} + {c^2}} \right)}}$
So the correct answer is option (B).

Note
For a spherical shell which has a constant surface charge density, then the potential due to it at a point greater than its radius will be the same as that of a point charge. For a point inside the shell, the potential will be constant but the electric field will be zero.