A surface with area $\vec S= 20\hat k$ is held in an electric field $\vec E= \hat i+2\hat j+5\hat k$. The electric flux linked with the surface is

100 units

The correct answer is Option (2) → 100 units

Given:

Electric field → $\vec{E} = \hat{i} + 2\hat{j} + 5\hat{k}$

Surface vector → $\vec{S} = 20\hat{k}$

Electric flux:

$\Phi = \vec{E} \cdot \vec{S}$

Substitute values:

$\Phi = (1\hat{i} + 2\hat{j} + 5\hat{k}) \cdot (20\hat{k})$

$\Phi = 1(0) + 2(0) + 5(20)$

$\Phi = 100$

Final Answer:

$\Phi = 100 \, \text{N·m}^2/\text{C}$