JEE PYQ: Electrostatics Question 20
Question 20 - 2020 (03 Sep Shift 2)
Concentric metallic hollow spheres of radii $R$ and $4R$ hold charges $Q_1$ and $Q_2$ respectively. Given that surface charge densities of the concentric spheres are equal, the potential difference $V(R) - V(4R)$ is:
(1) $\frac{3Q_1}{16\pi\varepsilon_0 R}$
(2) $\frac{3Q_2}{4\pi\varepsilon_0 R}$
(3) $\frac{Q_2}{4\pi\varepsilon_0 R}$
(4) $\frac{3Q_1}{4\pi\varepsilon_0 R}$
Show Answer
Answer: (1)
Solution
$\sigma_1 = \sigma_2$: $\frac{Q_1}{4\pi R^2} = \frac{Q_2}{4\pi(4R)^2}$, so $Q_2 = 16Q_1$. $V(R) = \frac{KQ_1}{R} + \frac{KQ_2}{4R}$ and $V(4R) = \frac{KQ_1}{4R} + \frac{KQ_2}{4R}$. $\Delta V = \frac{3KQ_1}{4R} = \frac{3Q_1}{16\pi\varepsilon_0 R}$.
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