The gravitational field due to a mass distribution is $I = k/x^{3}$ in the $x$-direction ($k$ is a constant). Taking the gravitational potential to be zero at infinity, its value at a distance $x/\sqrt{2}$ is

Question

Dr. Abubacker · Accepted Answer

(c) $dV = - Edx$
or $V = - \int_{\infty}^{x/\sqrt{2}}{Edx} = - \int_{\infty}^{x/\sqrt{2}}{kx^{- 3}dx = k/x^{2}}$

Q. The gravitational field due to a mass distribution is $I = k/x^{3}$ in the $x$-direction ($k$ is a constant). Taking the gravitational potential to be zero at infinity, its value at a distance $x/\sqrt{2}$ is

Explanation

Reviewed by Dr. Abubacker

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