Question

A copper ball of density 8.0 g/cc and 1 cm in diameter is immersed in oil of density 0.8 g/cc. The charge on the ball if it remains just suspended in oil in an electric field of intensity 600π V/m acting in the upward direction is_______.
Fill in the blank with the correct answer from the options given below. (Take g = 10 m/s²)

• A. 2 × 10⁻⁶C
• B. 1 × 10⁻⁶C
• C. 2 × 10⁻⁵C
• D. 1 × 10⁻⁵C

Answer 1

The Correct Option is C

Let's analyze the forces acting on the copper ball to determine the charge.

1. Given Values:

• Density of copper (ρ_c) = 8.0 g/cc = 8000 kg/m³
• Diameter of the ball (d) = 1 cm = 0.01 m
• Radius of the ball (r) = d/2 = 0.005 m
• Density of oil (ρ_o) = 0.8 g/cc = 800 kg/m³
• Electric field intensity (E) = 600π V/m (upward)
• Acceleration due to gravity (g) = 10 m/s²

2. Volume of the Ball (V):

V = (4/3)πr³

V = (4/3)π(0.005 m)³

V = (4/3)π(1.25 × 10^-7 m³)

V = (5/3)π × 10^-7 m³

3. Weight of the Ball (W):

W = mg = ρ_cVg

W = 8000 kg/m³ × (5/3)π × 10^-7 m³ × 10 m/s²

W = (40/3)π × 10^-3 N

4. Buoyant Force (F_b):

F_b = ρ_oVg

F_b = 800 kg/m³ × (5/3)π × 10^-7 m³ × 10 m/s²

F_b = (4/3)π × 10^-3 N

5. Electric Force (F_e):

F_e = qE

Where q is the charge on the ball.

6. Equilibrium Condition:

Since the ball is just suspended, the forces must balance:

W = F_b + F_e

(40/3)π × 10^-3 N = (4/3)π × 10^-3 N + q(600π V/m)

(40/3)π × 10^-3 - (4/3)π × 10^-3 = q(600π)

(36/3)π × 10^-3 = q(600π)

12π × 10^-3 = q(600π)

q = (12π × 10^-3) / (600π)

q = 12 × 10^-3 / 600

q = 2 × 10^-5 C

Therefore, the charge on the ball is 2 × 10^-5 C.

The correct answer is:

Option 2: 2 × 10^-5 C