The total charge enclosed in an incremental volume of 2× 10^-9 m^3 located at the origin is nC, if electric flux density of its field is found…

Electric flux density,

$$(\vec{D})=\frac{\text{charge}}{\text{Area}}\times \hat{r}=\frac{Q}{4\pi {r}^{2}}\hat{r}={\epsilon }_{0}(\frac{Q}{4\pi {\epsilon }_{0}{r}^{2}}\hat{r})$$(D)=Areacharge​×r^=4πr2Q​r^=ϵ0​(4πϵ0​r2Q​r^)

$$\Rightarrow \vec{E}=\frac{\vec{D}}{{\epsilon }_{0}}=\frac{{e}^{-x}siny\hat{i}-{e}^{-x}cos(\hat{j}+2z\hat{k)}}{{\epsilon }_{0}}$$⇒E=ϵ0​D​=ϵ0​e−xsinyi^−e−xcos(j^​+2zk)^​​

Also, by Gauss's law,

$$\frac{\rho }{{\epsilon }_{0}}=(\frac{\partial }{\partial x}\hat{i}+\frac{\partial }{\partial y}\hat{j}+\frac{\partial }{\partial z}\hat{k})\cdot \vec{E}=(\frac{\partial }{\partial x}\hat{i}+\frac{\partial }{\partial y}\hat{j}+\frac{\partial }{\partial z}\hat{k})\cdot \frac{\vec{D}}{{\epsilon }_{0}}$$ϵ0​ρ​=(∂x∂​i^+∂y∂​j^​+∂z∂​k^)⋅E=(∂x∂​i^+∂y∂​j^​+∂z∂​k^)⋅ϵ0​D​

$$\Rightarrow \rho =\frac{\partial }{\partial x}({e}^{-x}\mathrm{sin}y)+\frac{\partial }{\partial y}(-{e}^{-x}\mathrm{cos}y)+\frac{\partial }{\partial z}(2z)$$⇒ρ=∂x∂​(e−xsiny)+∂y∂​(−e−xcosy)+∂z∂​(2z)

$$\rho =-{e}^{-x}\mathrm{sin}y+{e}^{-x}\mathrm{sin}y+2$$ρ=−e−xsiny+e−xsiny+2

At origin $\rho =-{e}^{-0}\mathrm{sin}0+{e}^{-0}\mathrm{sin}0+2$ρ=−e−0sin0+e−0sin0+2

$$\rho =2C{m}^{-3}$$ρ=2Cm−3

Charge $=\rho \times$=ρ× volume $=2\times 2\times {10}^{-9}=4\times {10}^{-9}=$=2×2×10−9=4×10−9=

$$4\mathrm{nC}$$4nC