Question 779: Electrostats & Capacitance

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)

Accepted Answer

Step 1. Total charge on the sphere
The sphere has volume $V = \frac{4}{3} \pi r^3$.
So the total charge is
$ Q = \sigma \cdot V = \sigma \cdot \frac{4}{3} \pi r^3. $
Step 2. Apply GaussÃ¢ÂÂs law
For any $r_{gauss} \geq R$ (with $R = r$, the sphere radius), the sphere behaves like a point charge concentrated at the center.
GaussÃ¢ÂÂs law:
$ E \cdot 4\pi R^2 = \frac{Q}{\varepsilon_0}. $
Thus,
$ E = \frac{Q}{4\pi \varepsilon_0 R^2}. $
Step 3. Substitute $Q$
$ E = \frac{1}{4\pi \varepsilon_0 R^2}\,\left(	frac{4}{3}\pi R^3 \sigmaight) = \frac{\sigma R}{3\varepsilon_0}. $
Final Answer:
$ \boxed{	frac{\sigma r}{3\varepsilon_0}} $
So the correct option is
D
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Answer

zero

Answer

$\frac{5 \sigma r}{6 \varepsilon_0}$

Answer

$\frac{1}{4 \pi \varepsilon_0} \frac{\sigma}{\mathrm{r}}$

Answer

$\frac{\sigma \mathrm{r}}{3 \varepsilon_0}$

Electrostats & Capacitance

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)

Choose the correct answer: