A regular tetrhedron has four identical faces each an equilaterla triangle of side L. A charge +q is kept at one of the vertices. The magnitude of electric intensity due to this charge at thecentroid of the face opposite tio it is `(k=1//4 pe_(0))`

Options

A. `(2Kq)/(3L^(2))`
B. `(3Kq)/(2L^(2))`
C. `(3kq)/(L^(2))`
D. `(4kq)/(3L^(2))`

Explanation

To solve the problem of finding the electric intensity at the centroid of the face opposite to the vertex where a charge +q is placed in a regular tetrahedron, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Geometry of the Tetrahedron**:
- A regular tetrahedron has four vertices and four triangular faces. Each face is an equilateral triangle with side length \( L \).
- Let the vertices be labeled as \( A, B, C, \) and \( O \) (where \( O \) is the vertex with charge \( +q \)).

2. **Identify the Centroid of the Opposite Face**:
- The opposite face to vertex \( O \) is triangle \( ABC \). The centroid \( G \) of triangle \( ABC \) can be found using the formula for the centroid of a triangle, which is the average of the vertices' coordinates.

3. **Calculate the Distance from Charge to Centroid**:
- The distance from vertex \( O \) to the centroid \( G \) can be calculated using the Pythagorean theorem.
- The height \( h \) of the equilateral triangle \( ABC \) can be calculated as:
\[ h = \frac{\sqrt{3}}{2} L \]
- The distance from the centroid to the base (the line segment connecting the midpoints of the triangle) is \( \frac{h}{3} = \frac{L \sqrt{3}}{6} \).
- The vertical distance from \( O \) to the plane of triangle \( ABC \) is \( \frac{L}{\sqrt{3}} \) (the height from vertex \( O \) to the centroid of triangle \( ABC \)).

4. **Calculate the Total Distance from Charge to Centroid**:
- The total distance \( r \) from charge \( +q \) at vertex \( O \) to the centroid \( G \) can be found using:
\[ r = \sqrt{\left(\frac{L}{\sqrt{3}}\right)^2 + \left(\frac{L \sqrt{3}}{6}\right)^2} \]
- Simplifying this gives:
\[ r = \sqrt{\frac{L^2}{3} + \frac{L^2}{12}} = \sqrt{\frac{4L^2}{12} + \frac{L^2}{12}} = \sqrt{\frac{5L^2}{12}} = \frac{L \sqrt{5}}{2\sqrt{3}} \]

5. **Calculate the Electric Field Intensity**:
- The electric field intensity \( E \) due to a point charge is given by:
\[ E = \frac{k \cdot q}{r^2} \]
- Substituting \( r \):
\[ E = \frac{k \cdot q}{\left(\frac{L \sqrt{5}}{2\sqrt{3}}\right)^2} = \frac{k \cdot q}{\frac{5L^2}{12}} = \frac{12k \cdot q}{5L^2} \]

6. **Final Result**:
- The magnitude of the electric intensity at the centroid of the face opposite to the charge +q is:
\[ E = \frac{12kq}{5L^2} \]