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

Answer 1

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


$V=a{r}^2+b|E|=\frac{dV}{dr}=2ar\dots \left(1\right)Also,E=\frac{1}{4\pi {ϵ}_0}\frac{Q_{en}}{r^2}\dots \left(2\right)\left(1\right)\&\left(2\right)\Rightarrow -2ar=\frac{Q_{en}}{4\pi {ϵ}_0{r}^2}2a{r}^3=\frac{Q_{enc}}{4\pi {ϵ}_0}6a{r}^2=\frac{1}{4\pi {ϵ}_0}\frac{d{Q}_{en}}{dr}6a{r}^{2}dr=\frac{1}{4\pi {ϵ}_0}d{Q}_{enc}6a{r}^{2}dr=\frac{1}{4\pi {ϵ}_0}4\pi r^2\rho dr\therefore \rho 6a{ϵ}_0$