Electric Flux Through Shaded Area in Uniform Field

To find the electric flux through the given shaded surface, we can use two simple and equivalent approaches: using the vector area method or the projected area method.

Method 1: Using Area Vector

The vertices of the shaded surface, as given in the figure, are:

• $A = (0, 0, 0)$
• $B = (0, a, 0)$
• $C = (a, a, a)$
• $D = (a, 0, a)$

We can define two adjacent side vectors of the parallelogram $ABCD$:
$\vec{AB} = (0 - 0)\hat{i} + (a - 0)\hat{j} + (0 - 0)\hat{k} = a\hat{j}$
$\vec{AD} = (a - 0)\hat{i} + (0 - 0)\hat{j} + (a - 0)\hat{k} = a\hat{i} + a\hat{k}$

The vector area $\vec{S}$ of the planar surface is the cross product of these two side vectors:
$\vec{S} = \vec{AD} \times \vec{AB} = (a\hat{i} + a\hat{k}) \times (a\hat{j})$
$\vec{S} = a^2(\hat{i} \times \hat{j}) + a^2(\hat{k} \times \hat{j}) = a^2\hat{k} - a^2(-\hat{i}) = a^2\hat{i} + a^2\hat{k}$

The given uniform electric field is:
$\vec{E} = E_0 \hat{i}$

The electric flux $\Phi$ through the surface is given by the dot product of the electric field and the area vector:
$\Phi = \vec{E} \cdot \vec{S} = (E_0 \hat{i}) \cdot (a^2 \hat{i} + a^2 \hat{k})$
$\Phi = E_0 a^2 (\hat{i} \cdot \hat{i}) + E_0 a^2 (\hat{i} \cdot \hat{k})$
$\Phi = E_0 a^2 (1) + 0 = E_0 a^2$

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Method 2: Projected Area Method

Since the electric field $\vec{E} = E_0 \hat{i}$ is directed entirely along the $x$-axis, the flux through the surface is equal to the magnitude of the field multiplied by the area of the surface projected onto a plane perpendicular to the $x$-axis (the $yz$-plane).

1. Projecting the vertices onto the $yz$-plane gives:

   • $A' = (0, 0, 0)$
   • $B' = (0, a, 0)$
   • $C' = (0, a, a)$
   • $D' = (0, 0, a)$

2. The projection forms a square in the $yz$-plane of side length $a$.

3. The area of this projected region is:
   $A_{\text{proj}} = a \times a = a^2$

4. The flux through the shaded surface is therefore:
   $\Phi = E_0 \cdot A_{\text{proj}} = E_0 a^2$

Thus, the correct option is (C).