The gravitational field due to a mass distribution is E = (A)/(x^(2)) in x-direction. Here A is a constant. Taking the gravitaitonal potential to be zero at infinity , potential at x is

The gravitational field due to a mass distribution is E = (A)/(x^(2))

1.	The gravitational field due to a mass distribution is E = (A)/(x^(2)) in x-direction. Here A is a constant. Taking the gravitaitonal potential to be zero at infinity , potential at x is
Answer» `(2A)/(x)` `(2A)/(x)` `(A)/(x)` `(A)/(2X^(2))` ANSWER :C