In a region, the intensity of an electric field is given by $\overrightarrow{\mathrm{E}}=(2…
In a region, the intensity of an electric field is given by $\overrightarrow{\mathrm{E}}=(2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+\hat{\mathrm{k}}) \mathrm{NC}^{-1}$. The electric flux through a surface of area $10 \hat{\mathrm{i}} \mathrm{m}^2$ in the region is
$5 \mathrm{Nm}^2 \mathrm{C}^{-1}$
$10 \mathrm{Nm}^2 \mathrm{C}^{-1}$
$15 \mathrm{Nm}^2 \mathrm{C}^{-1}$
$20 \mathrm{Nm}^2 \mathrm{C}^{-1}$
Solution
An electric flux is given by
$\overrightarrow{\mathrm{E}}=(2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+\hat{\mathrm{k}}) \mathrm{N} / \mathrm{C}$
Surface of area, $\vec{A}=10 \hat{i} \mathrm{~m}^2$ Electric flux is given by
$\begin{aligned} & \phi=\overrightarrow{\mathrm{E}} \cdot \overrightarrow{\mathrm{A}} \\ & =(2 \hat{\mathrm{j}}+3 \hat{\mathrm{j}}+\hat{\mathrm{k}})(10 \hat{\mathrm{i}}) \\ & =20 \mathrm{~N} \mathrm{~m}^2 \mathrm{C}^{-1}\end{aligned}$