An electric field, $$\vec{E} = \frac{2\hat{i} + 6\hat{j} + 8\hat{k}}{\sqrt{6}}$$ passes through the surface of $$4 \text{ m}^2$$ 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 ______ Vm.

Find the electric flux through a surface of area 4 m$$^2$$.

$$\Phi = \vec{E} \cdot \vec{A} = \vec{E} \cdot (A\hat{n})$$

$$\vec{E} = \frac{2\hat{i} + 6\hat{j} + 8\hat{k}}{\sqrt{6}}$$, $$\hat{n} = \frac{2\hat{i} + \hat{j} + \hat{k}}{\sqrt{6}}$$, $$A = 4$$ m$$^2$$.

$$\vec{E} \cdot \hat{n} = \frac{1}{\sqrt{6}} \times \frac{1}{\sqrt{6}}(2\times2 + 6\times1 + 8\times1) = \frac{4+6+8}{6} = \frac{18}{6} = 3$$

$$\Phi = (\vec{E}\cdot\hat{n}) \times A = 3 \times 4 = 12 \text{ Vm}$$

The correct answer is 12.