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

To determine the charge on the copper ball required for it to remain suspended in the oil, we require the net force on the ball to be zero. The forces acting on the copper ball are:

1. Downward gravitational force: \( F_g = V \cdot \rho_c \cdot g \)
2. Upward buoyant force: \( F_b = V \cdot \rho_o \cdot g \)
3. Upward electrical force: \( F_e = qE \)

where:

• \(V\) is the volume of the ball
• \(\rho_c = 8.0 \, \text{g/cm}^3\) is the density of copper
• \(\rho_o = 0.8 \, \text{g/cm}^3\) is the density of oil
• \(g = 10 \, \text{m/s}^2\) is the acceleration due to gravity
• \(q\) is the charge on the ball
• \(E = 600\pi \, \text{V/m}\) is the electric field intensity

The volume \(V\) of the sphere is calculated as:

\(V = \frac{4}{3}\pi r^3\) with \(r = 0.5 \, \text{cm} = 0.005 \, \text{m}\).

Therefore,

\(V = \frac{4}{3}\pi (0.005)^3 = \frac{4}{3}\pi \times 1.25 \times 10^{-7} = \frac{5\pi}{3} \times 10^{-7} \, \text{m}^3\).

For the ball to be suspended, the total upward force must equal the downward force:

\((F_b + F_e) = F_g\).

Substituting the force expressions yields:

\(V \cdot \rho_o \cdot g + qE = V \cdot \rho_c \cdot g\).

Rearranging the equation to solve for \(q\):

\(q = \frac{V(\rho_c - \rho_o)g}{E}\).

Using the provided values:

• \(E = 600\pi\)
• \(\rho_c = 8000 \, \text{kg/m}^3\)
• \(\rho_o = 800 \, \text{kg/m}^3\)

The charge \(q\) is calculated as:

\(q = \frac{\left(\frac{5\pi}{3} \times 10^{-7}\right)(8000 - 800) \cdot 10}{600\pi} = \frac{\frac{5\pi}{3} \times 10^{-7} \cdot 7200 \cdot 10}{600\pi}\).

This simplifies to:

\(q = \frac{5 \times 7200 \times 10^{-7} \cdot 10}{3 \times 600} = \frac{5 \times 7200 \times 10^{-6}}{1800}\).

\(= \frac{5 \times 4}{10} \times 10^{-6} = 2 \times 10^{-5} \, \text{C}\).

The required charge on the copper ball is \(2 \times 10^{-5} \text{C}\).