A uniform electric field `Ehati` exists in a region containing the orign of co-ordinate system. The points `A(a,0,0),B(0,a,0)` and `C(0,0,a)` are on the co-ordinate axes. The magnitude of electric flux through the trianglular area ABC is

Options

A. `2Ea^(2)`
B. `Ea^(2)//2`
C. zero
D. `Ea^(2)`

Explanation

To find the magnitude of electric flux through the triangular area ABC in a uniform electric field \( \vec{E} \), we can follow these steps:

### Step 1: Understand the Geometry of the Triangle
The points A, B, and C are given as:
- \( A(a, 0, 0) \)
- \( B(0, a, 0) \)
- \( C(0, 0, a) \)

These points form a triangle in the coordinate system.

### Step 2: Find the Area of Triangle ABC
The area \( A \) of triangle ABC can be calculated using the formula for the area of a triangle formed by three points in 3D space. The formula is:
\[ A = \frac{1}{2} \times \text{base} \times \text{height} \]
In this case, we can take the base as the distance between points A and B, and the height as the perpendicular distance from point C to the line segment AB.

The distance AB can be calculated as:
\[ AB = \sqrt{(a - 0)^2 + (0 - a)^2} = \sqrt{a^2 + a^2} = a\sqrt{2} \]
The height from point C to line AB is \( \frac{a}{\sqrt{2}} \) (since the triangle is symmetrical).

Thus, the area \( A \) of triangle ABC is:
\[ A = \frac{1}{2} \times a\sqrt{2} \times \frac{a}{\sqrt{2}} = \frac{1}{2} \times a^2 = \frac{a^2}{2} \]

### Step 3: Determine the Electric Field Direction
Assuming the electric field \( \vec{E} \) is directed along the x-axis, we can express it as:
\[ \vec{E} = E \hat{i} \]

### Step 4: Calculate the Electric Flux
The electric flux \( \Phi_E \) through a surface is given by:
\[ \Phi_E = \vec{E} \cdot \vec{A} \]
where \( \vec{A} \) is the area vector of the triangle. The area vector is perpendicular to the surface and has a magnitude equal to the area of the triangle.

The area vector \( \vec{A} \) can be calculated as:
\[ \vec{A} = A \hat{n} \]
where \( \hat{n} \) is the unit normal vector to the surface. For triangle ABC, the normal vector can be determined using the right-hand rule, and it will not be aligned with the x-axis.

### Step 5: Calculate the Dot Product
The angle \( \theta \) between the electric field vector \( \vec{E} \) and the area vector \( \vec{A} \) can be determined. Since the area vector is not aligned with the x-axis, we need to find this angle.

Assuming the angle \( \theta \) is such that:
\[ \Phi_E = E \cdot A \cdot \cos(\theta) \]
Substituting the area we found:
\[ \Phi_E = E \cdot \frac{a^2}{2} \cdot \cos(\theta) \]

### Step 6: Final Expression for Electric Flux
The final expression for the electric flux through triangle ABC is:
\[ \Phi_E = \frac{E a^2}{2} \cos(\theta) \]