Question

The static electric field inside a dielectric medium with relative permittivity. ${ϵ}_r=2.25$, expressed in cylindrical coordinate system is given by the following expression
$E={\hat{a}}_{r}2+\left({\hat{a}}_{\phi }\frac{3}{r}\right)+{\hat{a}}_{z}6$
where ${\hat{a}}_r,{\hat{a}}_{\phi },{\hat{a}}_z$ are unit vectors along r,$\phi$ and z directions, respectively. If the above expression represents a valid electrostatic field inside the medium, then the volume charge density associated with this field in terms of free space permittivity, ${\epsilon }_0$ in SI unit is given by:

• A. $4{\epsilon }_0$
• B. $5{\epsilon }_0$
• C. $3{\epsilon }_0$
• D. $9{\epsilon }_0$

Answer 1

The correct option is D $9{\epsilon }_0$

Ans : (d)

$\overrightarrow{D}=ϵ\overrightarrow{E}={ϵ}_0{ϵ}_r\overrightarrow{E}$
$\overrightarrow{D}={ϵ}_{0}2.25(2r{\hat{a}}_r+\frac{3}{r}\hat{a}\varphi +6{\hat{a}}_Z)$
$\overrightarrow{D}=4.5{ϵ}_{0}r{\hat{a}}_r+\frac{6.75{ϵ}_0}{r}\hat{a}\varphi +13.5{ϵ}_0{\hat{a}}_Z$
${\rho }_V=▽.\overrightarrow{D}$
${\rho }_V=\frac{1}{r}\frac{\partial }{\partial r}\left(r{D}_r\right)+\frac{1}{r}\frac{\partial D_{\varphi }}{\partial \varphi }+\frac{\partial D_z}{\partial Z}$

$=\frac{1}{r}\frac{\partial }{\partial r}\left(4.5{ϵ}_0{r}^2\right)+0+0$
$=\frac{1}{r}\left(4.5{ϵ}_0\right(2r\left)\right)=9{ϵ}_0$