On the $x$-axis and a dsitance $x$ from the origin, the gravitational field due to a mass distribution
is given by $\frac{\mathrm{Ax}}{\left(\mathrm{x}^{2}+\mathrm{a}^{2}\right)^{3 / 2}}$ in the $\mathrm{x}$-direction. The
magnitude of gravitational potential on the $x$-axis at a distance $x$, taking its value to be zero at infinity, is :
Correct Option: 1,
Given $E_{G}=\frac{A x}{\left(x^{2}+a^{2}\right)^{3 / 2}}, V_{\infty}=0$
$\int_{V_{\infty}}^{V_{x}} d V=-\int_{\infty}^{x} \vec{E}_{G} \cdot \vec{d}_{x}$
$V_{x}-V_{\infty}=-\int_{\infty}^{x} \frac{A x}{\left(x^{2}+a^{2}\right)^{3 / 2}} d x$
put $x^{2}+a^{2}=z$
$2 x d x=d z$
$\mathrm{V}_{\mathrm{x}}-0=-\int_{\infty}^{\mathrm{x}} \frac{\mathrm{Adz}}{2(\mathrm{z})^{3 / 2}}=\left[\frac{\mathrm{A}}{\mathrm{z}^{1 / 2}}\right]_{\infty}^{\mathrm{x}}=\left[\frac{\mathrm{A}}{\left(\mathrm{x}^{2}+\mathrm{a}^{2}\right)^{1 / 2}}\right]_{\infty}^{\mathrm{x}}$
$\mathrm{V}_{\mathrm{x}}=\frac{\mathrm{A}}{\left(\mathrm{x}^{2}+\mathrm{a}^{2}\right)^{1 / 2}}-0=\frac{\mathrm{A}}{\left(\mathrm{x}^{2}+\mathrm{a}^{2}\right)^{1 / 2}}$
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