Question

Two small solid metal balls A and B of radii \(R\) and \(2R\) having charge densities 2 and 3 respectively are kept far apart. Find the charge densities on A and B after they are connected by a conducting wire.

Hint:

When two conductors are connected by a wire, the final charge densities are determined by the conservation of charge and the condition that the potentials on both spheres are equal. The radius of the spheres plays a crucial role in determining the final charge densities.

Answer 1

Let the charge densities on balls A and B before connecting by the wire be \(\sigma_A\) and \(\sigma_B\), and after connecting the wire, let the charge densities be \(\sigma_A'\) and \(\sigma_B'\). The initial charge densities are given as:
- On ball A: \( \sigma_A = 2 \)
- On ball B: \( \sigma_B = 3 \)
Step 1: Initial Charges on Balls The charge on a spherical ball is given by the formula: \[ Q = \sigma \cdot A = \sigma \cdot 4 \pi r^2 \] Where:
- \(r\) is the radius of the ball,
- \(\sigma\) is the surface charge density,
- \(A = 4 \pi r^2\) is the surface area of the ball. Thus, the initial charges on A and B are: For ball A: \[ Q_A = \sigma_A \cdot 4 \pi R^2 = 2 \cdot 4 \pi R^2 = 8 \pi R^2 \] For ball B: \[ Q_B = \sigma_B \cdot 4 \pi (2R)^2 = 3 \cdot 4 \pi (4R^2) = 48 \pi R^2 \]
Step 2: Total Charge and Conservation of Charge When the balls are connected by a conducting wire, they will share charge until they reach the same potential. Since the potentials on the balls must be the same, we use the formula for the potential on a spherical ball: \[ V = \frac{kQ}{r} \] Where:
- \(k\) is Coulomb’s constant,
- \(Q\) is the charge on the ball,
- \(r\) is the radius of the ball. Let the charges on balls A and B after the connection be \(Q_A'\) and \(Q_B'\). Since the potentials must be the same: \[ \frac{kQ_A'}{R} = \frac{kQ_B'}{2R} \] Simplifying: \[ Q_A' = \frac{1}{2} Q_B' \]
Step 3: Total Charge After Connection The total charge before the balls were connected is: \[ Q_{\text{total}} = Q_A + Q_B = 8 \pi R^2 + 48 \pi R^2 = 56 \pi R^2 \] Since charge is conserved, after the balls are connected, the total charge is the sum of the charges on the two balls: \[ Q_A' + Q_B' = 56 \pi R^2 \] Using the relation \( Q_A' = \frac{1}{2} Q_B' \), we substitute into the above equation: \[ \frac{1}{2} Q_B' + Q_B' = 56 \pi R^2 \] \[ \frac{3}{2} Q_B' = 56 \pi R^2 \] \[ Q_B' = \frac{2}{3} \cdot 56 \pi R^2 = 37.33 \pi R^2 \] Thus, the charge on ball B is \(Q_B' = 37.33 \pi R^2\). Now, using \( Q_A' = \frac{1}{2} Q_B' \): \[ Q_A' = \frac{1}{2} \cdot 37.33 \pi R^2 = 18.67 \pi R^2 \]
Step 4: Final Charge Densities Finally, the charge densities on A and B are given by: For ball A: \[ \sigma_A' = \frac{Q_A'}{4 \pi R^2} = \frac{18.67 \pi R^2}{4 \pi R^2} = 4.67 \] For ball B: \[ \sigma_B' = \frac{Q_B'}{4 \pi (2R)^2} = \frac{37.33 \pi R^2}{16 \pi R^2} = 2.33 \] Thus, the charge densities after the balls are connected by the wire are:
- \( \sigma_A' = 4.67 \)
- \( \sigma_B' = 2.33 \)