JEE PYQ: Electrostatics Question 18
Question 18 - 2020 (03 Sep Shift 1)
A charge $Q$ is distributed over two concentric conducting thin spherical shells radii $r$ and $R$ ($R > r$). If the surface charge densities on the two shells are equal, the electric potential at the common centre is:
(1) $\frac{1}{4\pi\varepsilon_0}\frac{(R+r)}{2(R^2+r^2)}Q$
(2) $\frac{1}{4\pi\varepsilon_0}\frac{(2R+r)}{(R^2+r^2)}Q$
(3) $\frac{1}{4\pi\varepsilon_0}\frac{(R+2r)}{2(R^2+r^2)}Q$
(4) $\frac{1}{4\pi\varepsilon_0}\frac{(R+r)}{(R^2+r^2)}Q$
Show Answer
Answer: (4)
Solution
Equal surface charge densities: $\sigma = \frac{Q_1}{4\pi r^2} = \frac{Q_2}{4\pi R^2}$. Total $Q = Q_1 + Q_2 = \sigma 4\pi(r^2 + R^2)$, so $\sigma = \frac{Q}{4\pi(r^2+R^2)}$. Potential at centre $V_c = \frac{KQ_1}{r} + \frac{KQ_2}{R} = K\sigma 4\pi(r + R) = \frac{1}{4\pi\varepsilon_0}\frac{(R+r)}{(R^2+r^2)}Q$.
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