Question

The gravitational potential at the centre of a square of side a when four point masses m each are kept at its vertices will be

• A. $4\sqrt{2}\frac{Gm}{a}$
• B. $-4\sqrt{2}\frac{Gm}{a}$
• C. $2\sqrt{2}\frac{Gm}{a}$
• D. $-2\sqrt{2}\frac{Gm}{a}$

Answer 1

The correct option is B $-4\sqrt{2}\frac{Gm}{a}$

$\begin{array}{c}AO=\frac{\sqrt{2}a}{2}=\frac{a}{\sqrt{2}}\\ V=\frac{-Gm}{r}=0\\ V_{net}=\frac{-4Gm}{r}=\frac{-4Gm}{\frac{a}{\sqrt{2}}}=\frac{-4\sqrt{2}Gm}{a}\end{array}$.

$\therefore$ Option $B$ is correct.