Question

If the energy stored in a spherical conductor having a charge of \( 12 \, \mu C \) is \( 6 \, \text{J} \), then the radius of the spherical conductor is:

• A. \( 10.8 \, \text{cm} \)
• B. \( 0.108 \, \text{cm} \)
• C. \( 1.08 \, \text{cm} \)
• D. \( 108 \, \text{cm} \)

Hint:

For a spherical conductor, the energy stored is inversely proportional to the radius. Use the formula $ U = \frac{Q^2}{8 \pi \epsilon_0 R} $ to relate energy, charge, and radius.

Answer 1

The Correct Option is A

Step 1: Known Information.
Charge on the spherical conductor: \( Q = 12 \, \mu C = 12 \times 10^{-6} \, C \)
Energy stored in the conductor: \( U = 6 \, J \)
The formula for the energy stored in a spherical conductor is: $$ U = \frac{Q^2}{8 \pi \epsilon_0 R} $$ where:
\( Q \) is the charge,
\( R \) is the radius of the conductor,
\( \epsilon_0 \) is the permittivity of free space (\( \epsilon_0 = 8.85 \times 10^{-12} \, F/m \)).
Step 2: Rearrange the Formula to Solve for \( R \).
Rearrange the energy formula to solve for \( R \):
$$ R = \frac{Q^2}{8 \pi \epsilon_0 U} $$ Step 3: Substitute Known Values.
Substitute the given values into the formula:
$$ Q = 12 \times 10^{-6} \, C, \quad \epsilon_0 = 8.85 \times 10^{-12} \, F/m, \quad U = 6 \, J $$ $$ R = \frac{(12 \times 10^{-6})^2}{8 \pi (8.85 \times 10^{-12})(6)} $$ Step 4: Simplify the Expression.
First, calculate \( Q^2 \): $$ Q^2 = (12 \times 10^{-6})^2 = 144 \times 10^{-12} = 1.44 \times 10^{-10} \, C^2 $$ Next, calculate the denominator:
$$ 8 \pi \epsilon_0 U = 8 \pi (8.85 \times 10^{-12})(6) $$ Approximate \( \pi \approx 3.1416 \):
$$ 8 \pi = 8 \times 3.1416 \approx 25.1328 $$ $$ 8 \pi \epsilon_0 U = 25.1328 \times (8.85 \times 10^{-12}) \times 6 $$ $$ 8 \pi \epsilon_0 U = 25.1328 \times 53.1 \times 10^{-12} \approx 1335.7 \times 10^{-12} = 1.3357 \times 10^{-9} $$ Now, calculate \( R \): $$ R = \frac{1.44 \times 10^{-10}}{1.3357 \times 10^{-9}} $$ Simplify: $$ R \approx \frac{1.44}{1.3357} \times 10^{-1} \approx 1.08 \times 10^{-1} \, m = 10.8 \, cm $$ Final Answer: \( \boxed{10.8 \, \text{cm}} \)