A point charge +Q is placed at the centroid of an equilateral triangle. When a second charge +Q is placed at a vertex of the triangle, the magnitude of the electrostatic force on the central charge is 8 N. The magnitude of the net force on the central charge when a third charge +Q is placed at another vertex of the triangle is

Options

A. zero
B. `4N`
C. `4(sqrt2) N`
D. `8 N`

Explanation

To solve the problem, we need to calculate the net electrostatic force acting on the central charge \( +Q \) when a third charge \( +Q \) is placed at another vertex of the equilateral triangle.

### Step-by-step Solution:

1. **Understanding the Configuration:**
- We have a point charge \( +Q \) at the centroid \( P \) of an equilateral triangle.
- A second charge \( +Q \) is placed at vertex \( A \).
- The electrostatic force on the charge at the centroid due to the charge at vertex \( A \) is given as \( 8 \, \text{N} \).

2. **Adding the Third Charge:**
- A third charge \( +Q \) is placed at another vertex \( B \) of the triangle.
- The distance from the centroid \( P \) to any vertex (including \( A \) and \( B \)) is the same due to the symmetry of the equilateral triangle.

3. **Forces Acting on the Central Charge:**
- The force \( F_{PA} \) due to the charge at vertex \( A \) is \( 8 \, \text{N} \) directed from \( P \) to \( A \).
- The force \( F_{PB} \) due to the charge at vertex \( B \) will also be \( 8 \, \text{N} \) directed from \( P \) to \( B \).

4. **Finding the Angle Between Forces:**
- Since the triangle is equilateral, the angle between the lines joining the centroid to the vertices \( A \) and \( B \) is \( 120^\circ \).

5. **Calculating the Net Force:**
- We can use the formula for the resultant of two forces \( F_1 \) and \( F_2 \) at an angle \( \theta \):
\[ F_{\text{net}} = \sqrt{F_1^2 + F_2^2 + 2 F_1 F_2 \cos \theta} \]
- Here, \( F_1 = 8 \, \text{N} \), \( F_2 = 8 \, \text{N} \), and \( \theta = 120^\circ \).
- The cosine of \( 120^\circ \) is \( -\frac{1}{2} \).

6. **Substituting Values:**
\[ F_{\text{net}} = \sqrt{8^2 + 8^2 + 2 \cdot 8 \cdot 8 \cdot \left(-\frac{1}{2}\right)} \]
\[ = \sqrt{64 + 64 - 64} \]
\[ = \sqrt{64} = 8 \, \text{N} \]

7. **Conclusion:**
- The magnitude of the net force on the central charge when a third charge \( +Q \) is placed at another vertex of the triangle is \( 8 \, \text{N} \).