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 \(\sigma_A\) and \(\sigma_B\) represent the initial charge densities on balls A and B, respectively, and \(\sigma_A'\) and \(\sigma_B'\) represent the charge densities after they are connected by a wire. The initial charge densities are provided as:
- Ball A: \( \sigma_A = 2 \)
- Ball B: \( \sigma_B = 3 \)
Step 1: Calculate Initial Charges The charge \(Q\) on a spherical object is determined by its surface charge density \(\sigma\) and surface area \(A\), using the formula \(Q = \sigma \cdot A\). For a sphere, the surface area is \(A = 4 \pi r^2\), where \(r\) is the radius. The initial charges on balls A and B are: For ball A, with radius \(R\): \[ Q_A = \sigma_A \cdot 4 \pi R^2 = 2 \cdot 4 \pi R^2 = 8 \pi R^2 \] For ball B, with radius \(2R\): \[ Q_B = \sigma_B \cdot 4 \pi (2R)^2 = 3 \cdot 4 \pi (4R^2) = 48 \pi R^2 \]
Step 2: Apply Charge Conservation and Potential Equality Upon connection with a conducting wire, the balls will redistribute charge until their electric potentials are equal. The potential \(V\) of a spherical object is given by \(V = \frac{kQ}{r}\), where \(k\) is Coulomb’s constant, \(Q\) is the charge, and \(r\) is the radius. Let \(Q_A'\) and \(Q_B'\) be the charges on balls A and B after connection. Since their potentials must equalize: \[ \frac{kQ_A'}{R} = \frac{kQ_B'}{2R} \] Simplifying this equation yields: \[ Q_A' = \frac{1}{2} Q_B' \]
Step 3: Determine Final Charges The total initial charge on the system is conserved. The total charge before connection is: \[ Q_{\text{total}} = Q_A + Q_B = 8 \pi R^2 + 48 \pi R^2 = 56 \pi R^2 \] After connection, the sum of the charges remains the same: \[ Q_A' + Q_B' = 56 \pi R^2 \] Substituting the relationship \( Q_A' = \frac{1}{2} Q_B' \) into this equation: \[ \frac{1}{2} Q_B' + Q_B' = 56 \pi R^2 \] \[ \frac{3}{2} Q_B' = 56 \pi R^2 \] Solving for \(Q_B'\): \[ Q_B' = \frac{2}{3} \cdot 56 \pi R^2 = 37.33 \pi R^2 \] The charge on ball B after connection is \(Q_B' = 37.33 \pi R^2\). Now, using \( Q_A' = \frac{1}{2} Q_B' \) to find the charge on ball A: \[ Q_A' = \frac{1}{2} \cdot 37.33 \pi R^2 = 18.67 \pi R^2 \]
Step 4: Calculate Final Charge Densities The final charge densities on balls A and B are calculated using their respective charges and surface areas. 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: \[ \sigma_B' = \frac{Q_B'}{4 \pi (2R)^2} = \frac{37.33 \pi R^2}{16 \pi R^2} = 2.33 \] Therefore, the charge densities after connecting the balls are:
- \( \sigma_A' = 4.67 \)
- \( \sigma_B' = 2.33 \)