Question

The gravitational field due to a mass distribution is $E=\frac{K}{x^3}$ in the $x-$direction, where $K$ is a constant. 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 $\frac{K}{2x^2}$

$dV=-Edx$
${\int }_0^{V}dV=-{\int }_{\infty }^x\frac{K}{x^3}dx$
$V=-{\left(\frac{K}{2x^2}\right)}_{\infty }^x=\frac{K}{2x^2}$