Question

find the potential energy of a system of four particles each of mass M placed at the vertices of a square of side l also obtain the potential at the centre of the square​

Answer 1

potential energy of a particle is given by,

P = GM₁M₂/r

considering particle 1 and 2

mass = M

distance = l

=> P₁ = GMM/l = GM²/l

similarly considering particle 1 and 4

mass = M

distance = l

=> P₂ = GMM/l = GM²/l

considering particle 1 and

mass = M

distance = l√2

=> P₃ = GMM/l√2 = GM²/l√2

Hence Net potential energy of the system,

P = P₁ + P₂ + P₃

P = (2 + 1/√2)GM²/l

Answer 2

Let square is ABCD.

Let mass= m at each point  

Let the side= a.  

Diagonal is= √2.

Potential energy for each pair of the masses. 

     U = - G m1 * m2 / d

Number of pairs:   4C2 = 6

Pairs are :  AB, AC, AD, BC, BD, CD.  

Masses distances in these pairs= a, √2 a, a, a, √2 a, a

Total Potential Energy = - G m * m * [ 1/a + 1/√2a  + 1/a + 1/a + 1/√2a + 1/a ]

           U = - G m²/a * [ 4 + √2 ]