Question

The electric field intensity on the surface of a charged sphere of radius R and volume charge density $\rho$ is:

• A. $\frac{3\epsilon_0}{R^2}$
• B. $\frac{1}{4\pi\epsilon_0} \frac{R^2}{\rho}$
• C. $\frac{R^2}{3\epsilon_0}$
• D. $\frac{\rho R}{3\epsilon_0}$

Hint:

Field inside grows linearly with radius; field outside follows inverse square law.

Answer 1

The Correct Option is D

Step 1: Know the charge spread.
A solid sphere of radius $R$ carries charge spread evenly through its whole volume. The volume charge density is $\rho$, meaning each unit of volume holds $\rho$ of charge.

Step 2: Find the total charge.
The volume of a sphere is $\dfrac{4}{3}\pi R^3$. So the total charge is $Q = \rho\cdot\dfrac{4}{3}\pi R^3$.

Step 3: Pick a Gaussian surface.
To find the field on the surface, imagine a sphere of radius $R$ around the charge. By symmetry the field points straight out and has the same size everywhere on this surface.

Step 4: Apply Gauss's law.
Gauss's law says the field times the surface area equals the enclosed charge divided by $\epsilon_0$. \[ E\,(4\pi R^2) = \frac{Q}{\epsilon_0} \]

Step 5: Put in the charge.
Replace $Q$ with our value: $E\,(4\pi R^2) = \dfrac{\rho\cdot\frac{4}{3}\pi R^3}{\epsilon_0}$.

Step 6: Solve for the field.
Divide both sides by $4\pi R^2$. The $4\pi$ cancels and one $R$ remains, giving $E = \dfrac{\rho R}{3\epsilon_0}$. \[ \boxed{\dfrac{\rho R}{3\epsilon_0}} \]