Question

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.

Answer 1

Answer :

consider a sqaure ABCD,and centre O

OA=1/2 AC =1/2 I root2 = I/root2

OA=OB=OC=OD=I/root2

Total potential energy = -2Gm2/I [ 2 + 1/root2]

= -5.414Gm2/I

Answer 2

Consider four masses each of mass m at the corners of a square of side $l$; See  Figure.

We have four mass pairs at distance $l$ and two diagonal pairs at distance $\sqrt{2}\: l$

   Hence,

        $W(r)=-4 \frac{Gm^2}{l} -2 \frac{Gm^2}{\sqrt{2}\:l}$

        $= \frac{2 Gm^2}{l} (2+\frac{1}{\sqrt{2}} = -5.41\frac{Gm^2}{l}$

The gravitational potential at the centre of the square $(r=\sqrt{2}l/2)$ is

$U(r)=-4\sqrt{2}\frac{Gm^2}{l}$