Question

The gravitational, field due to a mass distribution is $ E=\frac{k}{{{x}^{3}}} $ in the $ x $ - direction, where k is a constant. The value of gravitational potential at a distance $ x $ is [Taking gravitational potential to be zero at infinity]

• A. $ \frac{k}{x} $
• B. $ \frac{k}{{{x}^{3}}} $
• C. $ \frac{k}{2{{x}^{2}}} $
• D. $ \frac{k}{3{{x}^{3}}} $

Answer 1

Answer: (C)

Gravitation potential $=\int 1 dx$
Given, $1= E =\frac{ k }{ x ^{3}}$
$\therefore \int_{ x }^{\infty} \frac{ k }{ x ^{3}} dx$
$= k \left[\frac{ x ^{-3+1}}{-3+1}\right]_{ x }^{\infty}= k \left[\frac{-1}{2 x ^{2}}\right]_{ x }^{\infty}=\frac{ k }{2 x ^{2}}$