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The flux of the electric field $\overrightarrow{\mathrm{E}}=24 \hat{\mathrm{i}}+30 \hat{\mathrm{j}}+28 \hat{\mathrm{k}} \mathrm{NC}^{-1}$ through an area of $20 \mathrm{~m}^2$ on the yz plane is
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Verified Answer
The correct answer is:
$480 \mathrm{Nm}^2 \mathrm{C}^{-1}$
Electric field is given by
$$
\overrightarrow{\mathrm{E}}=24 \hat{\mathrm{i}}+30 \hat{\mathrm{j}}+28 \hat{\mathrm{k}}
$$
Area, $\vec{A}=20 \hat{i} \mathrm{~m}^2$
Electric flux, $\overrightarrow{\mathrm{D}}=\overrightarrow{\mathrm{E}} \cdot \overrightarrow{\mathrm{A}}$
$$
\begin{aligned}
& =(24 \hat{\mathrm{i}}+30 \hat{\mathrm{j}}+28 \hat{\mathrm{k}})(20 \hat{\mathrm{j}}) \\
& =480 \mathrm{~N} \mathrm{~m}^2 \mathrm{C}^{-1}
\end{aligned}
$$
$$
\overrightarrow{\mathrm{E}}=24 \hat{\mathrm{i}}+30 \hat{\mathrm{j}}+28 \hat{\mathrm{k}}
$$
Area, $\vec{A}=20 \hat{i} \mathrm{~m}^2$
Electric flux, $\overrightarrow{\mathrm{D}}=\overrightarrow{\mathrm{E}} \cdot \overrightarrow{\mathrm{A}}$
$$
\begin{aligned}
& =(24 \hat{\mathrm{i}}+30 \hat{\mathrm{j}}+28 \hat{\mathrm{k}})(20 \hat{\mathrm{j}}) \\
& =480 \mathrm{~N} \mathrm{~m}^2 \mathrm{C}^{-1}
\end{aligned}
$$
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