Question

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

• A. zero
• B. \(\frac{5\sigma r}{6\epsilon _{0}}\)
• C. \(\frac{1}{4\pi \epsilon _{0}}\frac{\sigma}{\mathrm{r}}\)
• D. \(\frac{\sigma \mathrm{r}}{3\epsilon _{0}}\)

Hint:

Apply Gauss's law with a spherical surface of radius r and charge Q = volume density times (4/3) pi r cubed.

Answer 1

The Correct Option is D

Step 1: Gauss's law
$E\,(4\pi r^2)=\frac{Q_{enc}}{\epsilon_0}$.

Step 2: Enclosed charge
$Q_{enc}=\sigma\cdot\frac{4}{3}\pi r^3$.

Step 3: Solve
$E=\frac{\sigma\frac{4}{3}\pi r^3}{4\pi r^2\epsilon_0}=\frac{\sigma r}{3\epsilon_0}$. Zero (option A) applies only to a hollow shell, so option (D) is correct.

Final Answer:
Option (D) is correct. \[ \boxed{\text{(D) }\dfrac{\sigma \mathrm{r}}{3\epsilon _{0}}} \]