Question

A uniformly charged conducting sphere of 2.4 m diameter has a surface charge density of 80.0 $ \mu C \, m^{-2} $. The charge on the sphere is nearly:

• A. \( 2.5 \times 10^{-3} \, \text{C} \)
• B. \( 1.45 \times 10^{-3} \, \text{C} \)
• C. \( 6.5 \times 10^{-3} \, \text{C} \)
• D. \( 0.15 \times 10^{-3} \, \text{C} \)

Hint:

For spherical conductors, the total charge is related to the surface charge density by \( Q = \sigma A \), where \( A \) is the surface area of the sphere and \( \sigma \) is the surface charge density.

Answer 1

The Correct Option is B

The charge on a conducting sphere can be calculated using the formula for surface charge density: \[ \sigma = \frac{Q}{A} \] where \( \sigma \) is the surface charge density and \( A \) is the surface area of the sphere. The surface area of a sphere is given by: \[ A = 4 \pi r^2 \] where \( r \) is the radius of the sphere. Given:
Diameter of the sphere \( d = 2.4 \, \text{m} \), so the radius is \( r = \frac{d}{2} = 1.2 \, \text{m} \), Surface charge density \( \sigma = 80.0 \, \mu C \, m^{-2} = 80.0 \times 10^{-6} \, C \, m^{-2} \). 
Now, calculate the charge: \[ A = 4 \pi r^2 = 4 \pi (1.2)^2 = 4 \pi (1.44) \approx 18.1 \, \text{m}^2 \] Using the formula \( \sigma = \frac{Q}{A} \), we can solve for \( Q \): \[ Q = \sigma \times A = 80.0 \times 10^{-6} \times 18.1 \approx 1.45 \times 10^{-3} \, \text{C} \] Conclusion: The charge on the sphere is \( 1.45 \times 10^{-3} \, \text{C} \). 
Final Answer: \[ \boxed{1.45 \times 10^{-3} \, \text{C}} \]