Question

A uniformly charged conducting sphere of diameter $3.5 \text{ cm}$ has a surface charge density of $20\mu\text{C m}^{-2}$. The total electric flux leaving the surface of the sphere is nearly}

• A. $57 \times 10^2 \text{ Vm}$
• B. $70 \times 10^2 \text{ Vm}$
• C. $87 \times 10^2 \text{ Vm}$
• D. $35 \times 10^3 \text{ Vm}$

Hint:

Flux depends only on the total enclosed charge, not the distribution. $\Phi = \sigma A / \varepsilon_0$.

Answer 1

The Correct Option is C

Step 1: Concept
Total electric flux $\Phi = Q/\varepsilon_0$ (Gauss's Law), where $Q = \sigma \cdot A$.

Step 2: Meaning
Surface area of sphere $A = \pi d^2$ (or $4\pi r^2$). $\sigma = 20 \times 10^{-6} \text{ C/m}^2$. $d = 0.035 \text{ m}$.

Step 3: Analysis
$\Phi = \frac{\sigma (\pi d^2)}{\varepsilon_0} = \frac{20 \times 10^{-6} \times 3.14 \times (0.035)^2}{8.85 \times 10^{-12}}$. $\Phi \approx \frac{20 \times 10^{-6} \times 0.00385}{8.85 \times 10^{-12}} \approx 8700$.

Step 4: Conclusion
The flux is approximately $87 \times 10^2 \text{ Vm}$ (or Wb equivalent unit). Final Answer: (C)