Question

A uniformly charged conducting sphere of \(0.2\,\text{m}\) diameter has a surface charge density of \(70\,\mu \text{C m}^{-2}\). The electric flux leaving the surface of the sphere is:

• A. \(9.9 \times 10^{5}\,\text{N C}^{-1}\text{m}^2\)
• B. \(9.9 \times 10^{6}\,\text{N C}^{-1}\text{m}^2\)
• C. \(8.9 \times 10^{5}\,\text{N C}^{-1}\text{m}^2\)
• D. \(8.9 \times 10^{6}\,\text{N C}^{-1}\text{m}^2\)

Hint:

Use Gauss’s law: \(\Phi = \frac{Q}{\varepsilon_0}\). Always compute total charge using surface area for surface charge density problems.

Answer 1

The Correct Option is A

Step 1: Use Gauss's law.
Electric flux through a closed surface is given by:
\[ \Phi = \frac{Q}{\varepsilon_0} \]

Step 2: Find total charge on the sphere.
\[ Q = \sigma \times 4\pi R^2 \]
Diameter \(= 0.2\,\text{m} \Rightarrow R = 0.1\,\text{m}\)
\[ \sigma = 70\,\mu \text{C m}^{-2} = 70 \times 10^{-6}\,\text{C m}^{-2} \]

Step 3: Substitute values.
\[ Q = 70 \times 10^{-6} \times 4\pi (0.1)^2 \]
\[ Q = 70 \times 10^{-6} \times 4\pi \times 0.01 \]
\[ Q = 70 \times 10^{-6} \times 0.04\pi \]
\[ Q = 2.8\pi \times 10^{-6} \]

Step 4: Apply Gauss's law.
\[ \Phi = \frac{2.8\pi \times 10^{-6}}{8.85 \times 10^{-12}} \]

Step 5: Simplify.
\[ \Phi \approx \frac{2.8 \times 3.14}{8.85} \times 10^{6} \]
\[ \Phi \approx 0.994 \times 10^{6} \]
\[ \Phi \approx 9.9 \times 10^{5}\,\text{N C}^{-1}\text{m}^2 \]

Step 6: Interpretation.
Electric flux depends only on total charge enclosed, not on shape or size of Gaussian surface.

Step 7: Final answer.
\[ \boxed{9.9 \times 10^{5}\,\text{N C}^{-1}\text{m}^2} \]