Electric flux through a surface area of 100m^2 lying in the xy plane (in Vm) if E = î + √2 ĵ + √3 k̂.

A. 100
B. 141.4
C. 173.2
D. 200

The electric flux ΦE\Phi_E through a surface is given by the formula:
ΦE=E⃗⋅A⃗\Phi_E = \vec{E} \cdot \vec{A}
Where:
• E⃗\vec{E} is the electric field vector,
• A⃗\vec{A} is the area vector, and
• ⋅\cdot denotes the dot product.
Given:
• The electric field vector is E⃗=i^+2j^+3k^ A/m\vec{E} = \hat{i} + \sqrt{2} \hat{j} + \sqrt{3} \hat{k} \, \text{A/m},
• The surface area is A=100 m2A = 100 \, \text{m}^2,
• The surface lies in the xyxy-plane, so the area vector is A⃗=100k^ m2\vec{A} = 100 \hat{k} \, \text{m}^2 (since the area vector is perpendicular to the surface and directed along the zz-axis).
Step 1: Dot Product of Electric Field and Area Vector
Now, we compute the dot product E⃗⋅A⃗\vec{E} \cdot \vec{A}:
E⃗⋅A⃗=(1i^+2j^+3k^)⋅(100k^)\vec{E} \cdot \vec{A} = (1 \hat{i} + \sqrt{2} \hat{j} + \sqrt{3} \hat{k}) \cdot (100 \hat{k})
Since the dot product of perpendicular vectors is zero, only the kk-component of E⃗\vec{E} will contribute to the flux:
E⃗⋅A⃗=100⋅3\vec{E} \cdot \vec{A} = 100 \cdot \sqrt{3} ΦE=1003 Vm\Phi_E = 100 \sqrt{3} \, \text{Vm}
Step 2: Simplifying the Result
3≈1.732\sqrt{3} \approx 1.732 ΦE=100×1.732=173.2 Vm\Phi_E = 100 \times 1.732 = 173.2 \, \text{Vm}
The electric flux through the surface is approximately 173.2 Vm, so the correct answer is C. 173.2.