In a region, the intensity of an electric field is given by $E=8 \hat{i}+8 \hat{j}+\hat{k}$ in $\mathrm{NC}^{-1}.$ The electric flux through a surface $S=10 \hat{i} m^2$ in the region is

$10 \mathrm{~Nm}^2 \mathrm{~C}^{-1}$

$8 \mathrm{~Nm}^2 \mathrm{~C}^{-1}$

JEE PHYSICS MCQ: ELECTRIC FIELDS AND CHARGES

1 Medium

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Application +4 | -1

In a region, the intensity of an electric field is given by

E=8 \hat{i}+8 \hat{j}+\hat{k}

\mathrm{NC}^{-1}.

The electric flux through a surface

S=10 \hat{i} m^2

in the region is

View answer and explanation

Correct option: B

80 \mathrm{~Nm}^2 \mathrm{~C}^{-1}

Option B (

80 \mathrm{~Nm}^2 \mathrm{~C}^{-1}

) is correct because electric flux is the dot product of the electric field and the area vector.

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r.

4N

\left( {{\varepsilon _r} = 8} \right)

N

5 \times {10^{ - 2}}