The gravitational field due to a mass distribution is E = K / x³ in the x - direction (K is a constant). Taking the gravitational potential to be zero at infinity, its value at a distance x is

The correct option is K / 2x².

Step-by-step Derivation:

The relation between the gravitational field intensity $E$ and the gravitational potential $V$ at a distance $x$ is given by the differential relation:
$E=-\frac{dV}{dx}$
This can be rewritten to solve for the potential $dV$ as:
$dV=-Edx$

We are given that the gravitational field is:
$E=\frac{K}{x^3}$
Substituting the expression for $E$ into the potential relation:
$dV=-\frac{K}{x^3}dx$

To find the potential $V\left(x\right)$ at a distance $x$, we integrate both sides from infinity (where the potential is defined to be zero, i.e., $V(\infty )=0$) to the distance $x$:
${\int }_0^{V\left(x\right)}dV=-{\int }_{\infty }^x\frac{K}{x^3}dx$

Integrating the right-hand side using the power rule for integration, $\int x^{-3}dx=\frac{x^{-2}}{-2}=-\frac{1}{2x^2}$:
$V\left(x\right)=-K{\left[-\frac{1}{2x^2}\right]}_{\infty }^x$
$V\left(x\right)=K{\left[\frac{1}{2x^2}\right]}_{\infty }^x$

Applying the upper and lower limits:
$V\left(x\right)=K\left[\frac{1}{2x^2}-\frac{1}{2(\infty {)}^2}\right]$
Since $\frac{1}{\infty }=0$, the equation simplifies to:
$V\left(x\right)=\frac{K}{2x^2}$