Question

Two isolated metallic spheres $S_1$ and $S_2$ of radii $1 \, \mathrm{cm}$ and $3 \, \mathrm{cm}$ respectively are charged such that both have the same charge density  \[ \sigma = \frac{2}{\pi} \times 10^{-9} \, \mathrm{C/m^2}. \] They are placed far away from each other and connected by a thin wire. Calculate the new charge on sphere $S_1$.
 

Hint:

When spheres are connected, charges redistribute such that their potentials are equal. Use the formula for potential and charge conservation to find the final charges.

Answer 1

Charge Redistribution Between Spheres
Charge on sphere \( S_1 \): \[ Q_1 = \sigma \times \text{Surface Area} = \left( \frac{2 \times 10^{-9}}{\pi} \right) \times 4\pi (1 \times 10^{-2})^2 = 8 \times 10^{-13} \, \text{C} \] Charge on sphere \( S_2 \): \[ Q_2 = \sigma \times \text{Surface Area} = \left( \frac{2 \times 10^{-9}}{\pi} \right) \times 4\pi (3 \times 10^{-2})^2 = 72 \times 10^{-13} \, \text{C} \] When the two spheres are connected by a thin wire, they acquire a common potential \( V \), and the charge remains conserved. Therefore: \[ Q_1 + Q_2 = Q_1' + Q_2' \] \[ = C_1 V + C_2 V \] Thus, \[ Q_1 + Q_2 = (C_1 + C_2) V \] The capacitances \(C_1\) and \(C_2\) are given by: \[ C_1 = 4\pi \epsilon_0 r_1 = \frac{1}{9} \times 10^{-11} \, \text{F} \] \[ C_2 = 4\pi \epsilon_0 r_2 = \frac{1}{3} \times 10^{-11} \, \text{F} \] Now, we calculate the common potential \( V \): \[ V = \frac{80 \times 10^{-13}}{\frac{1}{9} \times 10^{-11} + \frac{1}{3} \times 10^{-11}} = 1.8 \, \text{V} \] Substituting into \( Q_1' = C_1 V \), we get: \[ Q_1' = C_1 \times V = \frac{1}{9} \times 10^{-11} \times 1.8 = 2 \times 10^{-12} \, \text{C} \] Alternatively: Charge on sphere \( S_1 \): \[ Q_1 = \sigma \times \text{Surface Area} = \left( \frac{2 \times 10^{-9}}{\pi} \right) \times 4\pi (1 \times 10^{-2})^2 = 8 \times 10^{-13} \, \text{C} \] Charge on sphere \( S_2 \): \[ Q_2 = \sigma \times \text{Surface Area} = \left( \frac{2 \times 10^{-9}}{\pi} \right) \times 4\pi (3 \times 10^{-2})^2 = 72 \times 10^{-13} \, \text{C} \] When the spheres are connected by a thin wire, they acquire a common potential \( V \), and the charge remains conserved. Therefore: \[ Q_1 + Q_2 = Q_1' + Q_2' \] Additionally, we have the relation for the charge distribution: \[ \frac{Q_2'}{Q_1'} = \frac{r_2}{r_1} \] Solving this, we find: \[ Q_1' = 2 \times 10^{-12} \, \text{C} \]