Two metal spheres, one of radius $R$ and the other of radius $2R$ respectively have the same surface charge de

Question

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?

Accepted Answer

1.  **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 $Q_{	ext{total}} = Q_1 + Q_2 = 5Q_1 = 5\sigma(4\pi R^2)$.
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: $Q_1' + Q_2' = Q_{	ext{total}} \Rightarrow Q_1' + 2Q_1' = 5Q_1 \Rightarrow 3Q_1' = 5Q_1$.
    *   New charges: $Q_1' = \frac{5}{3}Q_1$ and $Q_2' = \frac{10}{3}Q_1$.
3.  **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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