An electric field is given by $(6 \hat{i}+5 \hat{j}+3 \hat{k}) \mathrm{N} / \mathrm{C}$. The electric flux through a surface area $30 \hat{i} \mathrm{~m}^2$ lying in YZ-plane (in SI unit) is :
Solution
<p>$$\begin{aligned}
& \overrightarrow{\mathrm{E}}=6 \hat{\mathrm{i}}+5 \hat{\mathrm{j}}+3 \hat{\mathrm{k}} \\
& \overrightarrow{\mathrm{A}}=30 \hat{\mathrm{i}} \\
& \phi=\overrightarrow{\mathrm{E}} \cdot \overrightarrow{\mathrm{A}} \\
& \phi=(6 \hat{\mathrm{i}}+5 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}) \cdot(30 \hat{\mathrm{i}}) \\
& \phi=6 \times 30=180
\end{aligned}$$</p>
About this question
Subject: Physics · Chapter: Electrostatics · Topic: Coulomb's Law
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