Question

Find the electric field in the sphere in first arrangement of figure.

• A. $\overrightarrow{E}=\frac{\rho r}{4\pi {\epsilon }_0}$
• B. $\overrightarrow{E}=\frac{\rho (\overrightarrow{r}-\overrightarrow{b})}{3\pi {\epsilon }_0}$
• C. $\overrightarrow{E}=\frac{\rho \overrightarrow{b}}{4\pi {\epsilon }_0}$
• D. none of these

Answer 1

Answer: (B)

$\overrightarrow{E}=\frac{\rho (\overrightarrow{r}-\overrightarrow{b})}{3\pi {\epsilon }_0}$

From above concept $\overrightarrow{E}=\frac{Kqr}{R^3}=\frac{1}{4\pi {ϵ}_{\circ }}\frac{4}{3a^3}\pi a^3\rho \overrightarrow{r^{\prime }}=\frac{\rho \overrightarrow{r^{\prime }}}{3{ϵ}_{\circ }}$ where $\overrightarrow{r^{\prime }}$ is position vector from center of sphere of radius $a$.

From Vector law of addition $\overrightarrow{r}=\overrightarrow{r^{\prime }}+\overrightarrow{b}$

$\Rightarrow \overrightarrow{r^{\prime }}=\overrightarrow{r}-\overrightarrow{b}$

$\therefore \overrightarrow{E}=\frac{\rho (\overrightarrow{r}-\overrightarrow{b})}{3{\epsilon }_0}$

The direction is radially out ward.