Q. The gravitational field due to a mass distribution is $1 = \frac{C}{x^{2}}$ in $x$ direction. Hence $C$ is constant. Taking the gravitational potential to be zero at infinity, potential at $x$ is
A) $\frac{2C}{x}$
B) $\frac{C}{x}$
C) $\frac{2C}{x^{2}}$
D) $\frac{C}{2x^{2}}$
Explanation
(b) $V = - \int_{\infty}^{x}{I\ dx = - \int_{\infty}^{x}{\frac{C}{x^{2}}\ dx = \frac{C}{x}}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.