Question

The electric field intensity on the surface of a charged solid sphere of radius $r$ and volume charge density $\rho$ is given by ($\varepsilon_0$ = permittivity of free space)

• A. zero
• B. $\frac{\rho r}{3\varepsilon_0}$
• C. $\frac{1}{4\pi\varepsilon_0}\frac{\rho}{r}$
• D. $\frac{5\rho}{6\pi\varepsilon_0}$

Hint:

Dimensional analysis can verify this immediately! Electric field units are derived from $\frac{\text{Charge}}{\varepsilon_0 \times \text{Area}}$. Since charge density $\rho$ is $\frac{\text{Charge}}{\text{Volume}}$, multiplying $\rho$ by a length parameter $r$ leaves you with exactly the correct $\frac{\text{Charge}}{\text{Area}}$ dimensions in the numerator, confirming the form $\frac{\rho r}{\varepsilon_0}$.

Answer 1

The Correct Option is B

Step 1: Use Gauss's law on the surface.
On the surface of a uniformly charged sphere the field behaves as if all the charge sat at the centre, so $E = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q}{r^2}$. We just need $q$ in terms of $\rho$.
Step 2: Express the total charge.
Charge equals density times volume: $q = \rho \cdot \dfrac{4}{3}\pi r^3$.
Step 3: Substitute into the field expression.
$E = \dfrac{1}{4\pi\varepsilon_0}\cdot\dfrac{\rho\,\tfrac{4}{3}\pi r^3}{r^2}$.
Step 4: Cancel the $4\pi$ factors.
The $4\pi$ in the denominator cancels with the $4\pi$ from the volume, leaving $E = \dfrac{\rho}{3\varepsilon_0}\cdot\dfrac{r^3}{r^2}$.
Step 5: Simplify the radius powers.
$\dfrac{r^3}{r^2} = r$, so $E = \dfrac{\rho r}{3\varepsilon_0}$.
Step 6: Conclude.
The surface field is $\dfrac{\rho r}{3\varepsilon_0}$. \[ \boxed{E = \frac{\rho r}{3\varepsilon_0}} \]