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

Answer 1

The correct option is C $\frac{k}{2x^2}$

Gravitation potential $=\int ldx$

Given, $l=E=\frac{k}{x^3}$

$\therefore$ ${\int }_x^{\infty }\frac{k}{x^3}dx$

$=k{\left[\frac{x^{-3+1}}{-3+1}\right]}_x^{\infty }=k{\left[\frac{-1}{2x^2}\right]}_x^{\infty }=\frac{k}{2x^2}$