Solved: PHYSICS Question for NEET

NEET PHYSICSMedium

Two metal spheres, one of radius $R$ and the other of radius $2R$ respectively have the same surface charge density $\sigma$ . They are brought in contact and separated. What will be the new surface charge densities on them?

$\sigma_1 = \frac{5}{6}\sigma, \sigma_2 = \frac{5}{6}\sigma$

$\sigma_1 = \frac{5}{2}\sigma, \sigma_2 = \frac{5}{6}\sigma$

$\sigma_1 = \frac{5}{2}\sigma, \sigma_2 = \frac{5}{3}\sigma$

$\sigma_1 = \frac{5}{3}\sigma, \sigma_2 = \frac{5}{6}\sigma$

Step-by-Step Solution

Initial Charge Calculation: Surface charge density $\sigma = \frac{Q}{A}$ [NCERT Class 12, Sec 1.12]. Charge on first sphere ( $R$ ): $Q_1 = \sigma (4\pi R^2)$ . Charge on second sphere ( $2R$ ): $Q_2 = \sigma (4\pi (2R)^2) = 4\sigma (4\pi R^2) = 4Q_1$ . Total charge $Q_{\text{total}} = Q_1 + Q_2 = 5Q_1 = 5\sigma(4\pi R^2)$ .
Redistribution of Charge: When brought in contact, charge flows until they reach a common potential $V$ . The potential of a conducting sphere is $V = \frac{1}{4\pi\epsilon_0}\frac{Q}{r}$ [NCERT Class 12, Sec 2.3]. Condition for equilibrium: $V_1' = V_2' \Rightarrow \frac{Q_1'}{R} = \frac{Q_2'}{2R} \Rightarrow Q_2' = 2Q_1'$ . Conservation of charge: $Q_1' + Q_2' = Q_{\text{total}} \Rightarrow Q_1' + 2Q_1' = 5Q_1 \Rightarrow 3Q_1' = 5Q_1$ . New charges: $Q_1' = \frac{5}{3}Q_1$ and $Q_2' = \frac{10}{3}Q_1$ .
Final Surface Charge Densities: $\sigma_1' = \frac{Q_1'}{4\pi R^2} = \frac{\frac{5}{3} \sigma (4\pi R^2)}{4\pi R^2} = \frac{5}{3}\sigma$ . $\sigma_2' = \frac{Q_2'}{4\pi (2R)^2} = \frac{\frac{10}{3} \sigma (4\pi R^2)}{4 (4\pi R^2)} = \frac{10}{12}\sigma = \frac{5}{6}\sigma$ .

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