Question

The gravitational field $E_g$ along $x-$axis, due to some distribution of mass is $E_g=\frac{k}{x^3}$. If the gravitational potential is taken as zero at infinity, then its value at a position $x$ is

• A. $\frac{k}{x}$
• B. $\frac{k}{2x}$
• C. $\frac{k}{x^2}$
• D. $\frac{k}{2x^2}$

Answer 1

The correct option is D $\frac{k}{2x^2}$

$E_g=\frac{k}{x^3}$

Since, $E_g=-\frac{d{V}_g}{dx}$

So $V_g=-{\int }_r^{\infty }{E}_{g}dx=-{\int }_r^{\infty }\frac{k}{x^3}dx$

$V_g=-(\frac{1}{\infty }-\frac{k}{2r^2})$

$V_g=\frac{k}{2r^2}$

putting $r=x$

$V_g=\frac{k}{2x^2}$