Question

Four point masses each of mass m are kept at four corners of a cube of side length ‘a’. Then

• A. The gravitational potential energy of system is $=-\frac{3\sqrt{2}G{m}^2}{a}$
• B. The gravitational potential energy of system is $=-\frac{4\sqrt{2}G{m}^2}{a}$
• C. The magnitude of force on one of the particle due to the remaining masses is equal to $\frac{\sqrt{6}G{m}^2}{a^2}$
• D. The magnitude of force on one of the particle due to others is equal to $\sqrt{\frac{3}{2}}\frac{G{m}^2}{a^2}$

Answer 1

Answer: (A), (D)

The magnitude of force on one of the particle due to others is equal to $\sqrt{\frac{3}{2}}\frac{G{m}^2}{a^2}$

Gravitational potential energy of the system
$U=-\frac{G{m}^2}{r}$ (for one face)
For all 6 faces, $U=-6\frac{G{m}^2}{r}$
Here, $r=a\sqrt{2}$
$U=-6\frac{G{m}^2}{a\sqrt{2}}=-\frac{3\sqrt{2}G{m}^2}{a}$
$F$ on $A=\frac{-G{m}^2}{(a\sqrt{2}{)}^2}$ due to one of the charges
$|F|=\frac{G{m}^2}{a^{3}2\sqrt{2}}\left[\sqrt{4a^2+4a^2+4a^2}\right]$
$=\sqrt{\frac{3}{2}}\frac{G{m}^2}{a^2}$