Question:
The electric field intensity on the surface of a charged sphere of radius R and volume charge density $\rho$ is:
The electric field intensity on the surface of a charged sphere of radius R and volume charge density $\rho$ is:
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Field inside grows linearly with radius; field outside follows inverse square law.
Updated On: Jun 10, 2026
- $\frac{3\epsilon_0}{R^2}$
- $\frac{1}{4\pi\epsilon_0} \frac{R^2}{\rho}$
- $\frac{R^2}{3\epsilon_0}$
- $\frac{\rho R}{3\epsilon_0}$
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The Correct Option is D
Solution and Explanation
Step 1: Concept
Gauss's Law: $\oint E \cdot dA = Q_{encl}/\epsilon_0$.
Step 2: Analysis
Total charge $Q = \rho \cdot (4/3)\pi R^3$. At the surface, $E(4\pi R^2) = Q/\epsilon_0 = (\rho \cdot (4/3)\pi R^3) / \epsilon_0$. $E = \frac{\rho \cdot (4/3)\pi R^3}{4\pi R^2 \epsilon_0} = \frac{\rho R}{3\epsilon_0}$.
Step 3: Conclusion
The electric field intensity is $\frac{\rho R}{3\epsilon_0}$.
Final Answer: (D)
Gauss's Law: $\oint E \cdot dA = Q_{encl}/\epsilon_0$.
Step 2: Analysis
Total charge $Q = \rho \cdot (4/3)\pi R^3$. At the surface, $E(4\pi R^2) = Q/\epsilon_0 = (\rho \cdot (4/3)\pi R^3) / \epsilon_0$. $E = \frac{\rho \cdot (4/3)\pi R^3}{4\pi R^2 \epsilon_0} = \frac{\rho R}{3\epsilon_0}$.
Step 3: Conclusion
The electric field intensity is $\frac{\rho R}{3\epsilon_0}$.
Final Answer: (D)
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