A thin wire of linear charge density `lambda` is bent in the form of a triangle ABC and an imaginary cube of side .a. is taken with vertex A at the centre of the top face and vertices B,C at the opposite edge centres of the bottom face. The total electric flux linked with the cube is

Options

A. `(lambdaa(sqrt3+1))/(in_(0))`
B. `(lambdaa(sqrt3+(1)/(2)))/(in_(0))`
C. `(lambdaa(sqrt5+1))/(in_(0))`
D. `(lambdaa(sqrt(5)+(1)/(2)))/(in_(0))`

Answer

To solve the problem of finding the total electric flux linked with the cube due to a thin wire of linear charge density `lambda` bent in the form of a triangle ABC, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Configuration**:
- We have a triangle ABC formed by a thin wire with linear charge density `lambda`.
- The cube has a side length `a`, with vertex A at the center of the top face and vertices B and C at the centers of the opposite edges of the bottom face.

2. **Using Gauss's Law**:
- According to Gauss's Law, the total electric flux (Φ) through a closed surface is given by:
\[ \Phi = \frac{Q_{\text{enclosed}}}{\epsilon_0} \]
- Here, \(Q_{\text{enclosed}}\) is the total charge enclosed by the Gaussian surface.

3. **Finding the Total Charge Enclosed**:
- The total charge \(Q_{\text{enclosed}}\) can be calculated as:
\[ Q_{\text{enclosed}} = \lambda \cdot L \]
- Where \(L\) is the total length of the wire that is enclosed within the cube.

4. **Calculating the Length of the Wire**:
- The triangle ABC has sides AB, AC, and BC.
- We know that BC = a (the side of the cube).
- To find AB and AC, we can use the Pythagorean theorem. The height from A to BC can be determined as follows:
- The distance from A to the midpoint of BC (let's call it P) is the height of the triangle.
- Since B and C are at the centers of the edges of the bottom face, the coordinates of B and C can be taken as \((-\frac{a}{2}, 0, 0)\) and \((\frac{a}{2}, 0, 0)\).
- The length AP = a (the height of the triangle).
- The distance from P to B (or C) is \(\frac{a}{2}\).

5. **Finding Lengths AB and AC**:
- Using the Pythagorean theorem:
\[ AB = AC = \sqrt{AP^2 + PB^2} = \sqrt{a^2 + \left(\frac{a}{2}\right)^2} = \sqrt{a^2 + \frac{a^2}{4}} = \sqrt{\frac{5a^2}{4}} = \frac{a\sqrt{5}}{2} \]
- Thus, the total length of the wire \(L\) is:
\[ L = AB + AC + BC = \frac{a\sqrt{5}}{2} + \frac{a\sqrt{5}}{2} + a = a\sqrt{5} + a = a(\sqrt{5} + 1) \]

6. **Calculating Total Charge Enclosed**:
- Now substituting back into the charge equation:
\[ Q_{\text{enclosed}} = \lambda \cdot a(\sqrt{5} + 1) \]

7. **Finding Total Electric Flux**:
- Finally, substituting \(Q_{\text{enclosed}}\) into Gauss's Law:
\[ \Phi = \frac{Q_{\text{enclosed}}}{\epsilon_0} = \frac{\lambda \cdot a(\sqrt{5} + 1)}{\epsilon_0} \]

### Final Answer:
Thus, the total electric flux linked with the cube is:
\[ \Phi = \frac{\lambda a (\sqrt{5} + 1)}{\epsilon_0} \]