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# The gravitational field due to mass distribution is $E= k/x^3$ in the x direction where k is a constant. Taking gravitational potential to be zero at infinity, its value at a distance x is

$(a)\;\frac{k}{x} \quad (b)\;\frac{k}{2x} \quad (c)\;\frac{k}{x^2}\quad (d)\;\frac{k}{2x^2}$

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A)
Gravitational Potential $=\int \limits_x^{\infty} E dx$
$\qquad=\int \large \frac{k}{x^3}$
$\qquad= -k \large\frac{x^{-2}}{2} \bigg|_x ^ {\infty}$
$\qquad=\large\frac{-k}{2x^2}$
magnitude is $\large\frac{k}{2x^2}$
Hence d is the correct answer.