Three charges, each of value `q`, are placed at the comers of an equilateral triangle. A fourth charge `Q` is placed at the centre of the triangle. If the value of `Q` for which charges remain stationary is `((-q)/(sqrt(n)))`, then find `n`.

Answer

To solve the problem, we need to find the value of \( n \) such that the charge \( Q \) placed at the center of an equilateral triangle formed by three charges \( q \) remains stationary.

### Step-by-Step Solution:

1. **Understanding the Configuration**:
- We have three charges \( q \) located at the corners of an equilateral triangle with side length \( a \).
- A fourth charge \( Q \) is placed at the center of the triangle.

2. **Finding the Distance from the Center**:
- The distance from the center of the triangle to any vertex (corner) is given by \( \frac{a}{\sqrt{3}} \).

3. **Calculating the Force on Charge \( Q \)**:
- Each charge \( q \) exerts a force on charge \( Q \). The force \( F \) between two point charges is given by Coulomb's law:
\[ F = k \frac{|qQ|}{r^2} \]
- Here, \( r = \frac{a}{\sqrt{3}} \), so the force due to one charge \( q \) on \( Q \) is:
\[ F = k \frac{qQ}{\left(\frac{a}{\sqrt{3}}\right)^2} = k \frac{qQ \cdot 3}{a^2} = \frac{3kqQ}{a^2} \]

4. **Resultant Force Calculation**:
- Since there are three identical charges \( q \) at the corners, the forces will have a symmetry. The net force \( F_{\text{net}} \) on charge \( Q \) due to the three charges can be calculated.
- The angle between any two forces due to the charges is \( 60^\circ \). Therefore, the resultant force can be calculated using the formula for the resultant of two forces:
\[ F_{\text{net}} = \sqrt{F^2 + F^2 + 2F \cdot F \cdot \cos(60^\circ)} = \sqrt{F^2 + F^2 + F^2} = \sqrt{3F^2} = F\sqrt{3} \]
- Hence, the net force acting on \( Q \) due to the three charges is:
\[ F_{\text{net}} = 3 \cdot \frac{3kqQ}{a^2} \cdot \frac{1}{2} = \frac{9kqQ}{2a^2} \]

5. **Condition for Equilibrium**:
- For charge \( Q \) to remain stationary, the net force acting on it must be zero. Therefore, we set the net force equal to zero:
\[ F_{\text{net}} = 0 \]
- This implies that the force due to the charge \( Q \) must balance out the forces from the three charges \( q \):
\[ \frac{9kqQ}{2a^2} = 0 \]

6. **Substituting the Value of \( Q \)**:
- Given that \( Q = -\frac{q}{\sqrt{n}} \), we substitute this into the equilibrium condition:
\[ \frac{9kq\left(-\frac{q}{\sqrt{n}}\right)}{2a^2} = 0 \]
- This leads to:
\[ -\frac{9kq^2}{2a^2\sqrt{n}} = 0 \]
- Since \( k \), \( q \), \( a \) are not zero, we conclude that:
\[ \sqrt{n} = 9 \implies n = 81 \]

### Final Answer:
Thus, the value of \( n \) is \( 81 \).