ઉગમબિંદુ પર રહેલા વિસ્તરતું કદ \(2 \times 10^{-9} \,{m}^{3}\) માં રહેલો વિદ્યુતભાર ...... \(nC\) હશે, જો તેના વિદ્યુતક્ષેત્રની વિદ્યુતફ્લક્સ ઘનતા \(D=e^{-x} \sin y \hat{i}-e^{-x} \cos y \hat{j}+2 z \hat{k}\, C / m^{2}\) હોય.

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

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

Also by Gauss's law

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

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

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

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

\(\rho=2 {C} / {m}^{3}\)

Charge \(=\rho \times\) volume \(=2 \times 2 \times 10^{-9}=4 \times 10^{-9}=4 {nC}\)