The gravitational field due to a mass distribution is `E=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 is

Options

A. `(K)/(x^(2))`
B. `(K)/(2x^(2))`
C. `(K)/(2x)`
D. `(K)/(x)`

Answer

B

Explanation

`E=-(dv)/(dx)" " therefore dV=-Edx` and `E=(K)/(x^(3))`
`therefore V= int_(x)^(oo)(K)/(x^(3))dx=K[(x^(-3+1))/(-3+1)]_(x)^(oo)`
`=-(K)/(2)[(1)/(x^(2))]_(x)^(oo)`
`V=-(K)/(2)[-(1)/(x^(2))]=(K)/(2x^(2))`