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

K/x

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K/2x

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Solution

The correct option is D

Gravitational potential = $\int E d x = \infty \int x \frac{K}{x^{3}} d x$
$= K (\frac{x^{- 3 + 1}}{- 3 + 1})_{x}^{\infty} = | \frac{- K}{2 x^{2}} |_{x}^{i n f t y} = \frac{K}{2 x^{2}}$

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Gravitational potential = $\int E d x = \infty \int x \frac{K}{x^{3}} d x$
$= K (\frac{x^{- 3 + 1}}{- 3 + 1})_{x}^{\infty} = | \frac{- K}{2 x^{2}} |_{x}^{i n f t y} = \frac{K}{2 x^{2}}$