JEE Main 2018 — Electrostatics Question with Solution

Let charge of shell A, B and C are Q_A, Q_B and Q_C respectively.

Potential of B shell will be due to charge Q_A, Q_B and Q_C.

Here charge Q_A is inside of the shell B and Q_B is on the surface of the shell B in both cases you have to take the radius of the shell B, while calculating potential of shell B.

Charge Q_C is outside of the shell B so take radius of shell C for calculating potential of shell B.

$$\therefore$$ V_B = V$${}_{Q_a}$$ + V$${}_{Q_b}$$ + V$${}_{Q_c}$$

= $$\frac{1}{4\pi {\epsilon }_0}$$ $$\left[\frac{4\pi a^2\left(+\sigma \right)}{b}+\frac{4\pi b^2\left(-\sigma \right)}{b}+\frac{4\pi c^2\left(+\sigma \right)}{c}\right]$$

= $$\frac{1}{4\pi {\epsilon }_0}$$ $$\times$$ 4$$\pi$$$$\sigma$$ $$\left[\frac{a^2}{b}-\frac{b^2}{b}+c\right]$$

= $$\frac{\sigma }{{\epsilon }_0}\left[\frac{a^2-b^2}{b}+c\right]$$