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The gravitational field due to a mass distribution is $E = \frac{K}{x^{3}}$ in the x-direction. (K is a constant). Taking the gravitational potential to be zero at infinity, its value at a distance x is

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

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

$\frac{K}{x}$

$\frac{K}{2 x}$

answer is D.

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Detailed Solution

$V = - \int_{\propto}^{x} E . dx = - \int_{\propto}^{x} \frac{k}{x^{3}} dx = \frac{k}{2 x^{2}}$

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