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

\frac{k}{x}

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\frac{k}{2 x}

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\frac{k}{x^{2}}

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\frac{k}{2 x^{2}}

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Solution

The correct option is D $\frac{k}{2 x^{2}}$
gravitational potential $V = - \int_{\infty}^{x} \to E . \to d x$
$V = - \int_{\infty}^{x} E d x$
$V = - \int_{\infty}^{x} \frac{k}{x^{3}} d x$
$V = - [\frac{k}{- 2 x^{2}} {∣ ∣ ∣}_{\infty}^{x}]$
$⟹ V = \frac{k}{2 x^{2}}$

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

The correct option is D $\frac{k}{2 x^{2}}$
gravitational potential $V = - \int_{\infty}^{x} \to E . \to d x$
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$⟹ V = \frac{k}{2 x^{2}}$