# Solve this following

Question:

The total charge enclosed in an incremental volume of $2 \times 10^{-9} \mathrm{~m}^{3}$ located at the origin is

$\mathrm{nC}$, if electric flux density of its field is found as $D=e^{-x} \sin y \hat{i}-e^{-x} \cos y \hat{j}+2 z \hat{k} \quad C / m^{2}$.

Solution:

Electric flux density

$(\overrightarrow{\mathrm{D}})=\frac{\text { charge }}{\text { Area }} \times \hat{\mathrm{r}}=\frac{\mathrm{Q}}{4 \pi \mathrm{r}^{2}} \hat{\mathrm{r}}=\epsilon_{0}\left(\frac{\mathrm{Q}}{4 \pi \epsilon_{0} \mathrm{r}^{2}} \hat{\mathrm{r}}\right)$

$\Rightarrow \vec{E}=\frac{\vec{D}}{\epsilon_{0}}=\frac{e^{-x} \sin y \hat{i}-e^{-x} \cos y \hat{j}+2 z \hat{k}}{\epsilon_{0}}$

Also by Gauss's law

$\frac{\rho}{\epsilon_{0}}=\left(\frac{\partial}{\partial \mathrm{x}} \hat{\mathrm{i}}+\frac{\partial}{\partial \mathrm{y}} \hat{\mathrm{j}}+\frac{\partial}{\partial \mathrm{z}} \hat{\mathrm{k}}\right) \cdot \overrightarrow{\mathrm{E}}=\left(\frac{\partial}{\partial \mathrm{x}} \hat{\mathrm{i}}+\frac{\partial}{\partial \mathrm{y}} \hat{\mathrm{j}}+\frac{\partial}{\partial \mathrm{z}} \hat{\mathrm{k}}\right) \cdot \frac{\overrightarrow{\mathrm{D}}}{\epsilon_{0}}$

$\Rightarrow \rho=\frac{\partial}{\partial x}\left(e^{-x} \sin y\right)+\frac{\partial}{\partial y}\left(-e^{-x} \cos y\right)+\frac{\partial}{\partial z}(2 z)$

$\rho=-e^{-x} \sin y+e^{-x} \sin y+2$

At origin $\rho=-e^{-0} \sin 0+e^{-0} \sin 0+2$

$\rho=2 C / m^{3}$

Charge $=\rho \times$ volume $=2 \times 2 \times 10^{-9}=4 \times 10^{-9}=$ $4 \mathrm{nC}$