NEET Electric Flux MCQs & PYQs

The area vector \( \vec{A} \) for a surface in the \( YZ \)-plane points along the \( \hat{i} \)-direction, with magnitude \( A = 20 \). Thus,
\[
\vec{A} = 20\hat{i}.
\]

The electric field is given by:
\[
\vec{E} = 5\hat{i} + 4\hat{j} + 9\hat{k}.
\]

Flux \( \Phi \) is:
\[
\Phi = \vec{E} \cdot \vec{A} = (5\hat{i} + 4\hat{j} + 9\hat{k}) \cdot 20\hat{i}.
\]

Only the \( \hat{i} \)-component contributes:
\[
\Phi = 5 \times 20 = 100 \, \text{units}.
\]

Specifying the total charge inside the surface is a sufficient condition for finding the electric flux \( \Phi \) through a closed surface.

According to Gauss's law:
\[
\Phi = \frac{q_{\text{enclosed}}}{\varepsilon_0}
\]

The flux depends only on the enclosed charge \( q_{\text{enclosed}} \), regardless of the surface's shape or the charge distribution.

By Gauss's law, the total flux through the cube is \(\Phi = \frac{q_{\text{enclosed}}}{ε_0}\). Since the charge is at the center, the flux through one of the six faces is \(\Phi_1 = \frac{\Phi}{6} = \frac{Q \times 10^{-6}}{6ε_0}\).