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

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

1.	The gravitational field due to a mass distributionis E= (A)/(x^2) in x-direction. Here A is a constant. Taking the gravitational potential to be zero at infinity, potential at x is
Answer» `(2A)/(X)` `(2A)/(x^3)` `(A)/(x)` `(A)/(2x^2)` ANSWER :C