Three equal charges `Q` are placed at the three vertices of an equilateral triangle. What should be the va,ue of a charge, that when placed at the centroid, reduces the interaction energy of the system to zero ?

Options

A. `(-Q)/2`
B. `(-Q)/3`
C. `(-Q)/(2sqrt(3))`
D. `(-Q)/(sqrt(3))`

Explanation

To solve the problem, we need to determine the value of a charge \( q \) that can be placed at the centroid of an equilateral triangle with three equal charges \( Q \) at its vertices, such that the total interaction energy of the system becomes zero.

### Step-by-Step Solution:

1. **Understanding the Configuration**:
- We have three equal charges \( Q \) located at the vertices of an equilateral triangle.
- Let the side length of the triangle be \( a \).
- The centroid of the triangle is the point where we will place the charge \( q \).

2. **Distance from Centroid to Vertices**:
- The distance from the centroid to each vertex of the triangle can be calculated using the formula:
\[ L = \frac{a}{\sqrt{3}} \]
- This is derived from the properties of an equilateral triangle.

3. **Calculating the Interaction Energy**:
- The interaction energy \( U \) of the system consists of two parts:
- The energy between the three charges at the vertices.
- The energy between the charge at the centroid and the three charges at the vertices.

4. **Energy Between Charges at Vertices**:
- The potential energy \( U_{vv} \) between the three charges at the vertices is given by:
\[ U_{vv} = \frac{kQ^2}{a} + \frac{kQ^2}{a} + \frac{kQ^2}{a} = 3 \frac{kQ^2}{a} \]

5. **Energy Between Centroid Charge and Vertex Charges**:
- The potential energy \( U_{vc} \) between the charge \( q \) at the centroid and each charge \( Q \) at the vertices is:
\[ U_{vc} = 3 \left( \frac{kQq}{L} \right) = 3 \left( \frac{kQq}{\frac{a}{\sqrt{3}}} \right) = \frac{3\sqrt{3}kQq}{a} \]

6. **Total Energy of the System**:
- The total energy \( U \) of the system is:
\[ U = U_{vv} + U_{vc} = 3 \frac{kQ^2}{a} + \frac{3\sqrt{3}kQq}{a} \]
- Simplifying, we get:
\[ U = \frac{3k}{a} \left( Q^2 + \sqrt{3}Qq \right) \]

7. **Setting Total Energy to Zero**:
- To find the charge \( q \) that makes the total energy zero, we set:
\[ \frac{3k}{a} \left( Q^2 + \sqrt{3}Qq \right) = 0 \]
- Since \( \frac{3k}{a} \) cannot be zero, we must have:
\[ Q^2 + \sqrt{3}Qq = 0 \]

8. **Solving for \( q \)**:
- Rearranging the equation:
\[ \sqrt{3}Qq = -Q^2 \]
- Dividing both sides by \( Q \) (assuming \( Q \neq 0 \)):
\[ q = -\frac{Q}{\sqrt{3}} \]

### Final Answer:
The value of the charge \( q \) that should be placed at the centroid to reduce the interaction energy of the system to zero is:
\[ q = -\frac{Q}{\sqrt{3}} \]