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Electrostatics question

2024 · 8 Apr · Shift 1 · Q82

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JEE Main
Physics
Electrostatics
2024 · 8 Apr · Shift 1 · Q82

Electrostatics question

2024 · 8 Apr · Shift 1 · Q82

JEE MainPhysicsElectrostaticsNumerical+4 / −1

An electric field,

\overrightarrow{\mathrm{E}}=\frac{2 \hat{i}+6 \hat{j}+8 \hat{k}}{\sqrt{6}}

passes through the surface of

4 \mathrm{~m}^2

area having unit vector

\hat{n}=\left(\frac{2 \hat{i}+\hat{j}+\hat{k}}{\sqrt{6}}\right)

. The electric flux for that surface is

\underline{\hspace{2cm}}\mathrm{Vm}

.

Numerical answer

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