Three positive charges of equal magnitude q are placed at the vertics of and equilatral triangle of side l. How can the system of charges be palced in equilibrium?

Options

A. by placing a charge `Q=-q//sqrt(3)` at the centroid of the triangle
B. by placing a charge `Q=q//sqrt(3)` at the centroid of the triangle
C. by placing a charge `Q=q` at the centroid of the triangle
D. by placing a charge `Q=-q` at the centroid of the triangle

Explanation

To solve the problem of placing three positive charges of equal magnitude \( q \) at the vertices of an equilateral triangle of side length \( l \) in equilibrium, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Configuration**:
- We have three positive charges \( +q \) located at the vertices \( A \), \( B \), and \( C \) of an equilateral triangle.
- The side length of the triangle is \( l \).

2. **Identifying Forces**:
- Each charge \( +q \) experiences repulsive forces due to the other two charges.
- The forces acting on any one charge can be calculated using Coulomb's Law.

3. **Calculating the Force Between Charges**:
- The force \( F \) between any two charges \( +q \) separated by a distance \( l \) is given by:
\[ F = k \frac{q^2}{l^2} \]
- Here, \( k \) is Coulomb's constant.

4. **Resultant Force on One Charge**:
- Consider the charge at vertex \( A \). The forces due to charges at \( B \) and \( C \) can be resolved into components.
- The angle between the lines connecting \( A \) to \( B \) and \( A \) to \( C \) is \( 60^\circ \).

5. **Resolving Forces**:
- The horizontal components of the forces from charges at \( B \) and \( C \) will add up, while the vertical components will cancel out.
- The resultant force \( F_R \) on charge \( A \) can be calculated as:
\[ F_R = 2F \cos(30^\circ) = 2 \left( k \frac{q^2}{l^2} \right) \left( \frac{\sqrt{3}}{2} \right) = k \frac{q^2 \sqrt{3}}{l^2} \]

6. **Introducing a Charge at the Centroid**:
- To achieve equilibrium, we need to introduce a charge \( -q' \) at the centroid \( O \) of the triangle.
- The centroid \( O \) is the point where the medians of the triangle intersect and is located at a distance \( \frac{l}{\sqrt{3}} \) from each vertex.

7. **Force on Charge at the Centroid**:
- The force \( F_O \) exerted on the charge \( -q' \) at the centroid due to each charge \( +q \) is:
\[ F_O = k \frac{q \cdot |q'|}{\left(\frac{l}{\sqrt{3}}\right)^2} = k \frac{q \cdot |q'| \cdot 3}{l^2} \]

8. **Setting Forces Equal for Equilibrium**:
- For the system to be in equilibrium, the resultant force \( F_R \) on charge \( A \) must equal the force \( F_O \) exerted on the charge at the centroid:
\[ k \frac{q^2 \sqrt{3}}{l^2} = k \frac{q \cdot |q'| \cdot 3}{l^2} \]
- Canceling \( k \) and \( l^2 \) gives:
\[ q^2 \sqrt{3} = 3 |q'| \]
- Solving for \( |q'| \):
\[ |q'| = \frac{q^2 \sqrt{3}}{3} \]

9. **Determining the Sign of the Charge**:
- Since we need a charge that attracts the positive charges, \( q' \) must be negative:
\[ q' = -\frac{q^2 \sqrt{3}}{3} \]

### Final Answer:
To place the system of charges in equilibrium, a charge of magnitude \( -\frac{q^2 \sqrt{3}}{3} \) should be placed at the centroid of the equilateral triangle.

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