Three point charges + q each are kept at the vertices of an equilateral triangle of side l. Determine the magnitude and sign of charge to be kept at the centroid so that charges at the vertices remain in equillibrium.

Options

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

Explanation

To solve the problem of determining the magnitude and sign of the charge to be kept at the centroid of an equilateral triangle such that the charges at the vertices remain in equilibrium, follow these steps:

### Step-by-Step Solution:

1. **Identify the given information:**
- Three charges \( +Q \) are placed at the vertices of an equilateral triangle with side length \( l \).

2. **Determine the distance from the centroid to a vertex:**
- For an equilateral triangle, the distance from the centroid to any vertex is \( \frac{l}{\sqrt{3}} \).

3. **Consider the forces acting on one charge at a vertex:**
- Each charge at a vertex experiences repulsive forces due to the other two charges at the vertices.
- The force between two charges \( +Q \) separated by a distance \( l \) is given by Coulomb's law:
\[ F = \frac{kQ^2}{l^2} \]
- Since the triangle is equilateral, the forces due to the other two charges will be equal in magnitude but will act along different directions.

4. **Resolve the forces into components:**
- Let’s consider the charge at vertex \( A \). The forces due to the charges at vertices \( B \) and \( C \) will have components along the line connecting \( A \) to the centroid and perpendicular to it.
- The horizontal components of the forces will cancel out due to symmetry, leaving only the vertical components to be considered.

5. **Calculate the resultant force due to the other two charges:**
- The vertical component of the force due to each charge is:
\[ F_{\text{vertical}} = F \cos(30^\circ) = \frac{kQ^2}{l^2} \cdot \frac{\sqrt{3}}{2} \]
- Since there are two such forces, the total vertical force is:
\[ F_{\text{total, vertical}} = 2 \cdot \frac{kQ^2}{l^2} \cdot \frac{\sqrt{3}}{2} = \frac{\sqrt{3}kQ^2}{l^2} \]

6. **Introduce the charge at the centroid:**
- Let the charge at the centroid be \( q \). This charge will exert an attractive force on the charge at vertex \( A \) because it must be negative to balance the repulsive forces.
- The force due to the charge \( q \) at the centroid on the charge at vertex \( A \) is:
\[ F_{\text{centroid}} = \frac{kQ|q|}{\left(\frac{l}{\sqrt{3}}\right)^2} = \frac{3kQ|q|}{l^2} \]

7. **Set up the equilibrium condition:**
- For the charge at vertex \( A \) to be in equilibrium, the attractive force due to the charge at the centroid must balance the resultant repulsive force:
\[ \frac{3kQ|q|}{l^2} = \frac{\sqrt{3}kQ^2}{l^2} \]
- Simplifying, we get:
\[ 3|q| = \sqrt{3}Q \]
\[ |q| = \frac{\sqrt{3}Q}{3} = \frac{Q}{\sqrt{3}} \]

8. **Determine the sign of the charge:**
- Since the charge at the centroid must be attractive to balance the repulsive forces, it must be negative:
\[ q = -\frac{Q}{\sqrt{3}} \]

### Final Answer:
The magnitude of the charge to be kept at the centroid is \( \frac{Q}{\sqrt{3}} \), and the sign of the charge is negative.