Question

The electric field intensity on the surface of a solid charged sphere of radius \( r \) and volume charge density \( \sigma \) is \( (\varepsilon_0 = \text{permittivity of free space}) \)

• A. zero
• B. \( \frac{\sigma}{\varepsilon_0} \)
• C. \( \frac{1}{4 \pi \varepsilon_0} \frac{\sigma}{r^2} \)
• D. \( \frac{1}{3 \varepsilon_0} \)

Hint:

Gauss's law is a powerful tool for calculating electric fields for symmetric charge distributions, such as spherical, cylindrical, or planar symmetries.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
We need the electric field on the surface of a uniformly charged solid sphere. Note that the text uses \( \sigma \) to denote volume charge density (usually denoted by \( \rho \)).
Step 2: Key Formula or Approach:
For a point on the surface (\( R = r \)):
\[ E = \frac{kQ}{r^2} \text{ where } Q = \text{volume} \times \text{density} \]
Step 3: Detailed Explanation:
Volume of the sphere \( V = \frac{4}{3} \pi r^3 \).
Total charge \( Q = V \cdot \sigma = \frac{4}{3} \pi r^3 \sigma \).
Substitute into the field formula:
\[ E = \frac{1}{4\pi\epsilon_0} \cdot \frac{\frac{4}{3} \pi r^3 \sigma}{r^2} \]
\[ E = \frac{\sigma \pi r^3}{3\pi\epsilon_0 r^2} = \frac{\sigma r}{3 \epsilon_0} \]
Step 4: Final Answer:
The electric field intensity is \( \frac{\sigma r}{3 \epsilon_0} \).