Question

The electric potential inside a charged sphere depends on the distance from its center as $V=a{r}^2+b$ where a and b are constants. The space charge distribution $\rho \left(r\right)$ inside the sphere is given by

• A. $\rho \left(r\right)=6a{\epsilon }_0$
• B. $\rho \left(r\right)=\frac{6a{\epsilon }_0}{r}$
• C. $\rho \left(r\right)=6a{\epsilon }_{0}r$
• D. $\rho \left(r\right)=36a{\epsilon }_0{r}^3$

Answer 1

The correct option is A $\rho \left(r\right)=6a{\epsilon }_0$



$V=a{r}^2+b$

$|E|=\frac{dV}{dr}=2ar------\left(1\right)$

Also, $E=\frac{1}{4\pi {\epsilon }_0}\frac{Q_{en}}{r^2}------\left(2\right)$

$\left(1\right)\&\left(2\right)\Rightarrow -2ar=\frac{Q_{en}}{4\pi {\epsilon }_0{r}^2}$

$2a{r}^3=\frac{Q_{enc}}{4\pi {\epsilon }_0}$

$6a{r}^2=\frac{1}{4\pi {\epsilon }_0}\frac{d{Q}_{en}}{dr}$

$6a{r}^{2}dr=\frac{1}{4\pi {\epsilon }_0}d{Q}_{enc}$

$6a{r}^{2}dr=\frac{1}{4\pi {\epsilon }_0}4\pi r^2\rho dr$

$\therefore \rho =6a{\epsilon }_0$