Three charges `q_(1),q_(2),q_(3)` each equal to q placed at the vertices of an equilateral triangle of side l . What will be the force on a charge Q placed at the centroid of triangle ?

Options

A. `4N`
B. `5N`
C. `10N`
D. Zero

Explanation

To find the force on a charge \( Q \) placed at the centroid of an equilateral triangle formed by three charges \( q_1, q_2, q_3 \) (each equal to \( q \)), we can follow these steps:

### Step 1: Understand the Configuration
We have three charges \( q_1, q_2, q_3 \) placed at the vertices of an equilateral triangle with side length \( l \). The centroid of the triangle is the point where the three medians intersect, and it is equidistant from all three vertices.

### Step 2: Calculate the Distance from the Centroid to Each Charge
For an equilateral triangle, the distance from the centroid to any vertex (let's denote it as \( AO \)) can be calculated as:
\[ AO = \frac{2}{3} \times \frac{l \sqrt{3}}{2} = \frac{l \sqrt{3}}{3} \]
This is derived from the property of centroids in an equilateral triangle.

### Step 3: Calculate the Force Exerted by Each Charge on Charge \( Q \)
Using Coulomb's law, the force \( F \) exerted by a charge \( q \) on a charge \( Q \) at a distance \( r \) is given by:
\[ F = k \frac{qQ}{r^2} \]
where \( k = \frac{1}{4\pi \epsilon_0} \).

Substituting \( r = AO = \frac{l \sqrt{3}}{3} \):
\[ F = k \frac{qQ}{\left(\frac{l \sqrt{3}}{3}\right)^2} = k \frac{qQ}{\frac{l^2 \cdot 3}{9}} = \frac{9kqQ}{3l^2} = \frac{3kqQ}{l^2} \]

### Step 4: Determine the Direction of Each Force
Since all three charges \( q_1, q_2, q_3 \) are positive, the forces \( F_1, F_2, F_3 \) acting on charge \( Q \) will be directed away from each charge towards the centroid.

### Step 5: Calculate the Net Force
The forces \( F_1, F_2, F_3 \) form angles of \( 120^\circ \) with each other due to the symmetry of the equilateral triangle.

Using the vector sum of forces:
- Let \( F_1 \) be directed along the negative x-axis.
- \( F_2 \) and \( F_3 \) will be at \( 120^\circ \) and \( 240^\circ \) respectively.

Using the cosine rule to find the resultant of \( F_2 \) and \( F_3 \):
\[ F_{\text{net}} = \sqrt{F_2^2 + F_3^2 + 2F_2F_3 \cos(120^\circ)} \]
Since \( F_2 = F_3 \), we can denote \( F_2 = F_3 = F \):
\[ F_{\text{net}} = \sqrt{F^2 + F^2 + 2F^2 \left(-\frac{1}{2}\right)} = \sqrt{2F^2 - F^2} = \sqrt{F^2} = F \]
Thus, the resultant force from \( F_2 \) and \( F_3 \) will be equal in magnitude to \( F_1 \) but in the opposite direction.

### Step 6: Conclusion
Since the forces \( F_1, F_2, F_3 \) are equal in magnitude and symmetrically arranged, they will cancel each other out. Therefore, the net force on charge \( Q \) at the centroid is:
\[ \text{Net Force} = 0 \]