Q 11-07-173JEE MainJEE Main 2020 (4 Sep, Shift 1)Medium
On the $x$-axis and at a distance $x$ from the origin, the gravitational field due to a mass distribution is given by $\dfrac{Ax}{(x^{2}+a^{2})^{3/2}}$ in the $x$-direction. The magnitude of the gravitational potential on the $x$-axis at a distance $x$, taking its value to be zero at infinity, is:
Answer: (A) $\dfrac{A}{(x^{2}+a^{2})^{1/2}}$
$$V(x) = -\int_\infty^{x} E\,dx = -\int_\infty^{x}\frac{Ax\,dx}{(x^{2}+a^{2})^{3/2}} = -\left[-\frac{A}{\sqrt{x^{2}+a^{2}}}\right]_\infty^{x} = \frac{A}{\sqrt{x^{2}+a^{2}}}$$
Magnitude: $\dfrac{A}{(x^{2}+a^{2})^{1/2}}$.
Solution by Sreeraj P, M.Sc Physics