An electric field is given by (6 hat{i}+5 hat{j}+3 hat{k}) N / C . electric flux through a surface a

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1. Electric Charges and Fields

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An electric field is given by $(6 \hat{i}+5 \hat{j}+3 \hat{k}) \ N / C$.

The electric flux through a surface area $30 \hat{\mathrm{i}}\; m^2$ lying in $YZ-$plane (in SI unit) is

A

$90$

B

$150$

C

$180$

D

$60$

(JEE MAIN-2024)

Solution

$ \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$

Standard 12

Physics

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