Find the potential energy of a system of four particles placed at the vertices of a square of side l. Also obtain the potential at the centre of the square.

Explanation

`implies`
As shown in figure, four masses each of mass m at the corners of a square of side l.
Four mass pairs at distance l and two diagonal pairs at distance `r = sqrt(2l)`
Potential energy  gained by each pair of particles,
`U_("ij")=-(Gm_(i)m_j)/(r_(ij))`
Total energy of system will the sum of potential energy of such pairs,
`:. U = -Gm^(2) [sum_(i=1)1/("rij")]" " [m_(i)=m,m_(j)=m]`
`:.U=-Gm^(2)[1/r_(12)+1/(r_(13))+1/(r_(14))+1/(r_23)+1/(r_24)+1/(r_34)]`
`:.U=-Gm^(2)[1/l+1/sqrt(2xxl)+1/l+1/l+1/(sqrt(2)xxl)+1/l]`
`:. U =-Gm^(2) [4/1+2/sqrt(2l)]`
`:. U =(-Gm^2)/l[4+sqrt2]`
Now , potential at the centre of square,
`phi=-G(m)/r`
`=-(Gm)/(l/sqrt2)" " [:. r = sqrt2l]`
`=-(sqrt(2)Gm)/l`
But there are four particles, each mass and distance from the centre are same , so
`phi=-(sqrt(2)Gm)/l`
But there are four particles, each mass and distance from the centre are same, so
`phi = - (4sqrt2Gm)/l`