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(A) The relationship between gravitational field $E_G$ and gravitational potential $V$ is given by $V(x) = -\int_{\infty}^{x} E_G \cdot dx$.Given $E_G

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$\frac{A}{(x^2+a^2)^{1/2}}$

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(A) The relationship between gravitational field $E_G$ and gravitational potential $V$ is given by $V(x) = -\int_{\infty}^{x} E_G \cdot dx$.Given $E_G = \frac{Ax}{(x^2+a^2)^{3/2}}$.Substituting this into the integral:$V(x) = -\int_{\infty}^{x} \frac{Ax}{(x^2+a^2)^{3/2}} dx$.Let $u = x^2 + a^2$,then $du = 2x dx$,or $x dx = \frac{du}{2}$.As $x 	o \infty$,$u 	o \infty$. As $x = x$,$u = x^2 + a^2$.$V(x) = -\int_{\infty}^{x^2+a^2} \frac{A}{u^{3/2}} \cdot \frac{du}{2} = -\frac{A}{2} \int_{\infty}^{x^2+a^2} u^{-3/2} du$.$V(x) = -\frac{A}{2} \left[ \frac{u^{-1/2}}{-1/2} ight]_{\infty}^{x^2+a^2} = A \left[ \frac{1}{\sqrt{u}} ight]_{\infty}^{x^2+a^2}$.$V(x) = A \left( \frac{1}{\sqrt{x^2+a^2}} - 0 ight) = \frac{A}{(x^2+a^2)^{1/2}}$.

On the $x$-axis and at a distance $x$ from the origin,the gravitational field due to a mass distribution is given by $\frac{Ax}{(x^2+a^2)^{3/2}}$ in the $x$-direction. The magnitude of gravitational potential on the $x$-axis at a distance $x$,taking its value to be zero at infinity,is

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