Electrostatics Part 3

Concept:For a spherical shell, potential inside and on the surface is constant and equals $\frac{kQ}{R}$, outside it is $\frac{kQ}{r}$. Use superposition to find total potential at any point.Explanation:Let $k = \frac{1}{4\pi\epsilon_0}$.For sphere A (radius $a$, charge $q_1$): point A is on its surface, inside B and C.Potential at A: $V_A = k\frac{q_1}{a} + k\frac{q_2}{b} + k\frac{q_3}{c} = \frac{1}{4\pi\epsilon_0}\left(\frac{q_1}{a}+\frac{q_2}{b}+\frac{q_3}{c}\right)$.For sphere B (radius $b$, charge $q_2$): point B is outside A, on surface of B, inside C.Potential at B: $V_B = k\frac{q_1}{b} + k\frac{q_2}{b} + k\frac{q_3}{c} = \frac{1}{4\pi\epsilon_0}\left(\frac{q_1+q_2}{b}+\frac{q_3}{c}\right)$.For sphere C (radius $c$, charge $q_3$): point C is outside A and B, on surface of C.Potential at C: $V_C = k\frac{q_1}{c} + k\frac{q_2}{c} + k\frac{q_3}{c} = \frac{1}{4\pi\epsilon_0}\left(\frac{q_1+q_2+q_3}{c}\right)$.Answer:$V_A = \frac{1}{4\pi\epsilon_0}\left(\frac{q_1}{a}+\frac{q_2}{b}+\frac{q_3}{c}\right)$, $V_B = \frac{1}{4\pi\epsilon_0}\left(\frac{q_1+q_2}{b}+\frac{q_3}{c}\right)$, $V_C = \frac{1}{4\pi\epsilon_0}\left(\frac{q_1+q_2+q_3}{c}\right)$.Hence, correct option is D.