The gravitational field due to a mass distribution is given by $E = \frac{- K}{x^{3}}$ in x-direction. Taking the gravitational potential to be zero at infinity, find its value at a distance ‘x’?

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$\frac{- K}{2 x^{2}}$

$\frac{- K}{2 x^{3}}$

$\frac{- K}{2 x^{4}}$

$\frac{- K}{2 x}$

answer is B.

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Detailed Solution

THE POTENTIAL AT A DISTANCE ‘X’ IS
$V = - \int E d x = \int_{\infty}^{x} \frac{K}{x^{3}} d x = \frac{- K}{2 x^{2}} |_{\infty}^{x} ∴ V = \frac{- K}{2 x^{2}}$

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