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 solve this problem, we want to find the charge on the copper ball such that it remains suspended in the oil. This means that the net force on the ball is zero. The forces acting on the copper ball are: 

1. Gravitational force acting downward: \( F_g = V \cdot \rho_c \cdot g \)
2. Buoyant force acting upward: \( F_b = V \cdot \rho_o \cdot g \)
3. Electrical force acting upward: \( F_e = qE \)

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 given by:

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

Thus,

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

The condition for suspension is that the net upward force equals the downward force:

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

Substituting the expressions for each force, we have:

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

Rearranging for \(q\),

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

Substitute the given values:

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

Therefore,

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

This simplifies to:

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

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

Thus, the correct answer is: \(2 \times 10^{-5} \text{C}\).