Question

A spherical metal ball of density \( \rho \) and radius \( r \) is immersed in a liquid of density \( \sigma \). When an electric field is applied in the upward direction the metal ball remains just suspended in the liqui(D) Then the expression for the charge on the metal ball is:

• A. \( q = \frac{4\pi r^3 \rho}{\varepsilon} \)
• B. \( q = \frac{4\pi r^3 (\rho - \sigma)}{\varepsilon} \)
• C. \( q = \frac{4\pi r^3 \sigma}{\varepsilon} \)
• D. \( q = \frac{4\pi r^3 (\rho - \sigma)}{\varepsilon} \)

Hint:

For objects just suspended in a liquid under an electric field, use the balance between the electrostatic force and the buoyant force to derive the expression for the charge.

Answer 1

The Correct Option is B

When a spherical metal ball is immersed in a liquid, and an electric field is applied in the upward direction, the ball just remains suspended when the upward electrostatic force balances the downward buoyant force. 1. The buoyant force \( F_B \) is equal to the weight of the liquid displaced by the ball, which is given by: \[ F_B = \text{Weight of displaced liquid} = \rho_{\text{liquid}} \cdot V_{\text{ball}} \cdot g \] where: - \( V_{\text{ball}} \) is the volume of the ball - \( \rho_{\text{liquid}} \) is the density of the liquid Since the ball has radius \( r \), the volume of the ball is: \[ V_{\text{ball}} = \frac{4}{3} \pi r^3 \] 2. The electrostatic force \( F_E \) on the charged ball due to the electric field \( E \) is given by: \[ F_E = qE \] where \( q \) is the charge on the ball. 3. For the ball to remain suspended, the upward electrostatic force must balance the downward buoyant force. Thus: \[ qE = \text{Buoyant force} \] 4. The buoyant force is also given by the difference in densities between the ball and the liquid, i.e., \( \rho - \sigma \). Substituting the expression for buoyant force, we get: \[ qE = \frac{4}{3} \pi r^3 (\rho - \sigma) g \] Since the electric field is related to the charge by \( E = \frac{q}{\varepsilon} \), we substitute this into the equation: \[ q \cdot \frac{q}{\varepsilon} = \frac{4}{3} \pi r^3 (\rho - \sigma) g \] Solving for \( q \), we get: \[ q = \frac{4 \pi r^3 (\rho - \sigma)}{\varepsilon} \] Thus, the correct expression for the charge on the metal ball is: \[ q = \frac{4\pi r^3 (\rho - \sigma)}{\varepsilon} \]