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

Explanation

Refer to Fig. `AC = BD = l sqrt(2)`

`OA = OB = OC = OD = (l sqrt(2))/(2) = (l)/(sqrt(2))`
Using superposition principle, total potentila energy of the system of four particles placed at the vertices `A, B, C` and `D` of a square is
`U = ((-Gm xx m)/(AB)) + ((-Gm xx m)/(BC) + (-Gm xx m)/(AC))`
`+ ((-Gm xxm)/(AD) + (- Gm xx m)/(BD) + (-Gm xx m)/(CD))`
`= (-4 Gm^(2))/(l) - (2 Gm^(2))/(l sqrt(2))`
`= (- 2 Gm^(2))/(l) [2 + (1)/(sqrt(2))] = (-5.414 Gm^(2))/(l)`
Total gravitational potential at the centre `O `of the square is
`V = (-Gm xx 4)/(OA) = (-4 Gm)/(l//sqrt(2)) = - 4 sqrt(2) (Gm)/(l)`