Four particles each of mass m are placed at the vertices of a square of side `l`. the potential at the centre of square is

Options

A. `-2sqrt(2) (Gm)/l`
B. `-3sqrt(2) (Gm)/l`
C. `-2(Gm)/l`
D. `-4sqrt(2) (Gm)/l`

Explanation

To find the gravitational potential at the center of a square formed by four particles each of mass \( m \) placed at the vertices, we can follow these steps:

### Step 1: Understand the Setup
We have four masses \( m \) located at the corners of a square with side length \( l \). We need to calculate the gravitational potential at the center of this square.

### Step 2: Calculate the Distance from the Center to a Vertex
The center of the square is equidistant from all four vertices. To find this distance, we can use the Pythagorean theorem. The distance from the center to any vertex (let's say vertex A) can be calculated as follows:

- The coordinates of the vertices can be taken as \( (0, 0) \), \( (l, 0) \), \( (0, l) \), and \( (l, l) \).
- The center of the square is at \( \left(\frac{l}{2}, \frac{l}{2}\right) \).
- The distance \( d \) from the center to a vertex (for example, vertex A at \( (0, 0) \)) is given by:
\[ d = \sqrt{\left(\frac{l}{2} - 0\right)^2 + \left(\frac{l}{2} - 0\right)^2} = \sqrt{\left(\frac{l}{2}\right)^2 + \left(\frac{l}{2}\right)^2} = \sqrt{\frac{l^2}{4} + \frac{l^2}{4}} = \sqrt{\frac{l^2}{2}} = \frac{l}{\sqrt{2}} \]

### Step 3: Write the Formula for Gravitational Potential
The gravitational potential \( V \) due to a mass \( m \) at a distance \( r \) is given by:
\[ V = -\frac{Gm}{r} \]
where \( G \) is the gravitational constant.

### Step 4: Calculate the Total Potential at the Center
Since potential is a scalar quantity, we can simply add the potentials due to each mass at the center. The distance from the center to each mass is \( \frac{l}{\sqrt{2}} \).

Thus, the total potential \( V_{total} \) at the center due to all four masses is:
\[ V_{total} = V_A + V_B + V_C + V_D = -\frac{Gm}{\frac{l}{\sqrt{2}}} - \frac{Gm}{\frac{l}{\sqrt{2}}} - \frac{Gm}{\frac{l}{\sqrt{2}}} - \frac{Gm}{\frac{l}{\sqrt{2}}} \]
\[ V_{total} = -4 \cdot \frac{Gm}{\frac{l}{\sqrt{2}}} = -\frac{4Gm\sqrt{2}}{l} \]

### Final Answer
The potential at the center of the square is:
\[ V_{total} = -\frac{4Gm\sqrt{2}}{l} \]

---