Question

The gravitational field in due to some mass distribution is $E=k/x^3$, in $X-$direction. Taking the gravitational potential to be zero at infinity, its value at a distance $x$ is:

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

Answer 1

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

We know that-

$E=-\frac{dV}{dx}$

$⟹dV=-Edx$

$⟹V=-{\int }_{\infty }^x\frac{k}{x^3}dx$

$⟹V=-{\int }_{\infty }^{x}k{x}^{-3}dx$

$⟹V=-{\left[\frac{k{x}^{-3+1}}{-3+1}\right]}_{\infty }^x$

$⟹V=\frac{k}{2}{\left[\frac{1}{x^2}\right]}_{\infty }^x=\frac{k}{2}[\frac{1}{x^2}-\frac{1}{\infty }]$

$⟹V=\frac{k}{2x^2}$

Answer-(D)