Step 1: Understanding the Concept:
For the ball to be suspended, the net force on it must be zero.
Upward forces: Electric force (\( qE \)) and Buoyant force (\( F_B \)).
Downward force: Weight (\( mg \)).
Step 2: Key Formula or Approach:
Equilibrium equation: \( qE + V \rho_{oil} g = V \rho_{ball} g \)
Rearranging: \( qE = V g (\rho_{ball} - \rho_{oil}) \)
Where \( V \) is the volume of the ball.
Step 3: Detailed Explanation:
1. Volume calculation:
Diameter \( D = 1 \, \text{cm} \implies \text{Radius } R = 0.5 \, \text{cm} = 0.5 \times 10^{-2} \, \text{m} \).
\[ V = \frac{4}{3} \pi R^3 = \frac{4}{3} \pi (0.5 \times 10^{-2})^3 = \frac{4}{3} \pi \times 0.125 \times 10^{-6} \, \text{m}^3 \]
2. Densities in SI units:
\( \rho_{ball} = 8.0 \, \text{g/cc} = 8000 \, \text{kg/m}^3 \)
\( \rho_{oil} = 0.8 \, \text{g/cc} = 800 \, \text{kg/m}^3 \)
3. Substituting into the balance equation:
\[ q \times 600\pi = \left( \frac{4}{3} \pi \times 0.125 \times 10^{-6} \right) \times 10 \times (8000 - 800) \]
Cancel \( \pi \) and simplify:
\[ 600 q = \frac{4}{3} \times 0.125 \times 10^{-5} \times 7200 \]
\[ 600 q = \left( \frac{4 \times 7200}{3} \right) \times 0.125 \times 10^{-5} \]
\[ 600 q = (4 \times 2400) \times 0.125 \times 10^{-5} \]
\[ 600 q = 9600 \times 0.125 \times 10^{-5} \]
\[ 600 q = 1200 \times 10^{-5} \]
\[ q = \frac{1200}{600} \times 10^{-5} = 2 \times 10^{-5} \, \text{C} \]
Step 4: Final Answer:
The charge on the ball is $2 \times 10^{-5}$ C.