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

Explanation

Four particle , each of mass m, are placed at the four corners of a square ABCD of side l as shown in the Fig.1.12.

Here AO=CO =BO =DO=`l/sqrt(2)`
We know that the gravitational potential energy associated with two particles of masses `m_1 and m_2` at a
distance r is `-G(m_1m_2)/r`.
Here, masses at the four corners are at a distance l from one another and the two diagonal pairs are at a distance `l/sqrt(2)+l/(sqrt(2) =sqrt(2)l.`
So, the gravitational potential energy of the whole system,
`E_p =4(-(Gm^2)/l)+2(-G(m^2)/(sqrt(2)l))=-2G(m^2)/l(2+1/sqrt(2))`
=`-5.41(Gm^2)/l`
Gravitational potential at the centre (O) of the square is
`V=4(-(Gm)/(l/sqrt(2)))=-4sqrt(2)(Gm)/l`