On the x-axis and at a distance $x$ from the origin, the gravitational field due to a mass distribution is given by
On the $x$-axis and at a distance $x$ from the origin, the gravitational field due to a mass distribution is given by
$\frac{A x}{\left(x^{2}+a^{2}\right)^{3 / 2}}$ 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 :
Correct Option: 1
(1) Given: Gravitational field,
$E_{G}=\frac{A x}{\left(x^{2}+a^{2}\right)^{3 / 2}}, V_{\infty}=0$
$\int_{V_{\infty}}^{V_{x}} d V=-\int_{\infty}^{x} \vec{E}_{G} \cdot \vec{d}_{x}$
$\Rightarrow V_{x}-V_{\infty}=-\int_{\infty}^{x} \frac{A x}{\left(x^{2}+a^{2}\right)^{3 / 2}} d x$
$\therefore V_{x}=\frac{A}{\left(x^{2}+a^{2}\right)^{1 / 2}}-0=\frac{A}{\left(x^{2}+a^{2}\right)^{1 / 2}}$
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