Question

The gravitational field due to a mass distribution is given by $E=\frac{K}{r^3}$ in $\text{x-}$direction. Taking the gravitational potential to be zero at infinity, find its value at a distance $x$

• A. $\frac{K}{2x^2}$
• B. $\frac{K}{3x^2}$
• C. $-\frac{2K}{3x^2}$
• D. $-\frac{3K}{2x^2}$

Answer 1

The correct option is A $\frac{K}{2x^2}$

The change in potential while going from the initial to final positions in a gravitational field $E$ is given by
$$dV=-{\int }_i^f\overrightarrow{E}.\overrightarrow{dr}$$ Since, the change is only in the $\text{x-}$direction, so the potential at distance $x$ is ,

$V=-{\int }_{\infty }^{x}Edr$

$\Rightarrow V=-{\int }_{\infty }^x\frac{K}{r^3}dr$

$\Rightarrow V=-K{\int }_{\infty }^x\frac{1}{r^3}dr$

$\Rightarrow V=K{\left[\frac{1}{2r^2}\right]}_{\infty }^x$

$\Rightarrow V=\frac{K}{2}[\frac{1}{x^2}-\frac{1}{{\infty }^2}]$

$\therefore V=\frac{K}{2x^2}$

Hence, option (a) is the correct answer.