CGPET 2021 Solved Paper

Consider a square of side l and four particles each of mass m‌kg placed at the vertices P,Q,R and S.

In ∆SRQ,
SQ=√(SR)2+(RQ)2⇒SQ=l√2
Since, SO=RO=QO=PO=

l√2

2

=

1

√2

Now, potential at the centre due to one mass
=−

Gm

l √2

=−√2

Gm

l

Therefore, potential at the centre due to all four masses (each of m),
Vnet=4×(−√2

Gm

l

)⇒Vnet−4√2

Gm

l