Question

Given that the surface charge density on a sphere is $200\, \mu \text{C/m^2$, what is the electric field at the surface of the sphere?}

• A. $ 1.13 \times 10^4 \, \text{N/C} $
• B. $ 2.26 \times 10^4 \, \text{N/C} $
• C. $ 2.26 \times 10^6 \, \text{N/C} $
• D. $ 1.13 \times 10^6 \, \text{N/C} $

Hint:

Use the formula $ E = \frac{\sigma}{\varepsilon_0} $ to calculate the electric field due to surface charge density.

Answer 1

The Correct Option is B

Step 1: Recall the formula for electric field at the surface of a charged sphere
For a uniformly charged conducting sphere, the electric field just outside its surface is given by: \[ E = \frac{\sigma}{\varepsilon_0} \] where: $\sigma$ is the surface charge density,
$\varepsilon_0$ is the vacuum permittivity, with value $\varepsilon_0 = 8.85 \times 10^{-12} \, \text{C}^2/\text{N} \cdot \text{m}^2$. \vspace{0.3cm} Step 2: Use the given values
Given:
$\sigma = 200 \, \mu\text{C/m}^2 = 200 \times 10^{-6} \, \text{C/m}^2$ Substitute into the formula:
\[ E = \frac{200 \times 10^{-6}}{8.85 \times 10^{-12}} = \frac{2 \times 10^{-4}}{8.85 \times 10^{-12}} \] \[ E \approx 2.26 \times 10^{7} \, \text{N/C} \] \vspace{0.3cm} Step 3: Conclusion
The electric field at the surface of the sphere is: \[ {2.26 \times 10^{7} \, \text{N/C}} \]