Question

A copper ball of density $8.0\text{ g/cc}$ and $1\text{ cm}$ in diameter is immersed in oil of density $0.8\text{ g/cc}$. The charge on the ball if it remains just suspended in oil in an electric field of intensity $600\pi\text{ V/m}$ acting in the upward direction is __________. Fill in the blank with the correct answer from the options given below. (Take $g = 10\text{ m/s}^2$)

• A. $2 \times 10^{-6}\text{ C}$
• B. $2 \times 10^{-5}\text{ C}$
• C. $1 \times 10^{-5}\text{ C}$
• D. $1 \times 10^{-6}\text{ C}$

Hint:

When converting density from $\text{g/cm}^3$ to $\text{kg/m}^3$, simply multiply by $1000$. For example, $8.0\text{ g/cc} = 8000\text{ kg/m}^3$.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
For the ball to remain suspended, the net force must be zero. The forces acting are: 1. Weight ($mg$) - Downward 2. Buoyant Force ($F_B$) - Upward 3. Electric Force ($qE$) - Upward

Step 2: Identifying the Formula and Values:
The equilibrium equation is: $qE + F_B = mg \implies qE = mg - F_B$. This can be written using density ($\rho$ for ball, $\sigma$ for oil) and volume ($V$): \[ qE = V\rho g - V\sigma g = Vg(\rho - \sigma) \] Values:
• $r = 0.5\text{ cm} = 0.5 \times 10^{-2}\text{ m}$
• $V = \frac{4}{3}\pi r^3 = \frac{4}{3}\pi (0.5 \times 10^{-2})^3 = \frac{4}{3}\pi (0.125 \times 10^{-6})\text{ m}^3$
• $\rho - \sigma = (8.0 - 0.8)\text{ g/cc} = 7.2\text{ g/cm}^3 = 7200\text{ kg/m}^3$

Step 3: Calculation:
\[ q(600\pi) = \left[\frac{4}{3}\pi \times 0.125 \times 10^{-6}\right] \times 10 \times 7200 \] \[ 600q = \frac{4}{3} \times 0.125 \times 10^{-5} \times 7200 \] \[ 600q = 4 \times 0.125 \times 10^{-5} \times 2400 \] \[ q = \frac{0.5 \times 10^{-5} \times 2400}{600} = 0.5 \times 10^{-5} \times 4 = 2 \times 10^{-5}\text{ C} \]

Step 4: Final Answer:
The charge on the ball is $2 \times 10^{-5}\text{ C}$.