 # On the x-axis and at a distance $x$ from the origin, the gravitational field due to a mass distribution is given by Question:

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 :

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

2. (2) $\frac{A}{\left(x^{2}+a^{2}\right)^{3 / 2}}$

3. (3) $A\left(x^{2}+a^{2}\right)^{1 / 2}$

4. (4) $A\left(x^{2}+a^{2}\right)^{3 / 2}$

Correct Option: 1

Solution:

(1) Given: Gravitational field,

$E_{G}=\frac{A x}{\left(x^{2}+a^{2}\right)^{3 / 2}}, V_{\infty}=0$

$\int_{V_{\infty}}^{V_{x}} d V=-\int_{\infty}^{x} \vec{E}_{G} \cdot \vec{d}_{x}$

$\Rightarrow V_{x}-V_{\infty}=-\int_{\infty}^{x} \frac{A x}{\left(x^{2}+a^{2}\right)^{3 / 2}} d x$

$\therefore V_{x}=\frac{A}{\left(x^{2}+a^{2}\right)^{1 / 2}}-0=\frac{A}{\left(x^{2}+a^{2}\right)^{1 / 2}}$