The total charge enclosed in an incremental v | vedclass

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Question

Electric flux density

$(\vec{D})=\frac{	ext { charge }}{	ext { Area }} 	imes \hat{r}=\frac{Q}{4 \pi r^{2}} \hat{r}=\epsilon_{0}\left(\frac{Q}{4

Vedclass · Accepted Answer

$4$

Vedclass · Answer

Electric flux density

$(\vec{D})=\frac{	ext { charge }}{	ext { Area }} 	imes \hat{r}=\frac{Q}{4 \pi r^{2}} \hat{r}=\epsilon_{0}\left(\frac{Q}{4 \pi \epsilon_{0} r^{2}} \hat{r}ight)$

$\Rightarrow \vec{E}=\frac{\vec{D}}{\epsilon_{0}}=\frac{e^{-x} \sin y \hat{i}-e^{-x} \cos y \hat{j}+2 z \hat{k}}{\epsilon_{0}}$

Also by Gauss's law

$\frac{ho}{\epsilon_{0}}=\left(\frac{\partial}{\partial x} \hat{i}+\frac{\partial}{\partial y} \hat{j}+\frac{\partial}{\partial z} \hat{k}ight) \cdot \vec{E}=\left(\frac{\partial}{\partial x} \hat{i}+\frac{\partial}{\partial y} \hat{j}+\frac{\partial}{\partial z} \hat{k}ight) \cdot \frac{\vec{D}}{\epsilon_{0}}$

$\Rightarrow ho=\frac{\partial}{\partial x}\left(e^{-x} \sin yight)+\frac{\partial}{\partial y}\left(-e^{-x} \cos yight)+\frac{\partial}{\partial z}(2 z)$

$ho=-e^{-x} \sin y+e^{-x} \sin y+2$

At origin $ho=-e^{-0} \sin 0+e^{-0} \sin 0+2$

$ho=2 {C} / {m}^{3}$

Charge $=ho 	imes$ volume $=2 	imes 2 	imes 10^{-9}=4 	imes 10^{-9}=4 {nC}$

The total charge enclosed in an incremental volume of $2 \times 10^{-9} \,{m}^{3}$ located at the origin is ...... $nC,$ if electric flux density of its field is found as $D=e^{-x} \sin y \hat{i}-e^{-x} \cos y \hat{j}+2 z \hat{k}\, C / m^{2}$

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