On the x -axis and at a distance x from the origin, the gravitational field due to a mass distribution is given by Ax ( x ^ 2 + a ^ 2 ) ^ 3 / 2 in the x -direction. The magnitude of the gravitational potential on the x

Correct answer: A. A ( x ^ 2 + a ^ 2 ) ^ 1 / 2 V _ x = - ∫ _ ∞ ^ x Ax ( A ^ 2 + x ^ 2 ) ^ 3 / 2 (- dx) V _ x = - A A ^ 2 + x ^ 2

JEE Main 20204 SeptemberMorning Shift

JEE Main 2020 · 4 September · Morning Shift · Physics · Gravitation

On the x -axis and at a distance x from the origin, the...

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}{{\left({x}^{2} + {a}^{2}\right)}^{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:

A. $\dfrac{A}{{\left({x}^{2} + {a}^{2}\right)}^{1 / 2}}$

${V}_{x} = - \mathop{∫}_{\infty}^{x}\dfrac{Ax}{{\left({A}^{2} + {x}^{2}\right)}^{3 / 2}}\left(- dx\right)$

${V}_{x} = - \dfrac{A}{\sqrt{{A}^{2} + {x}^{2}}}$