Question

On the $x-$axis 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

• A. $\frac{A}{(x^2+a^2{)}^{1/2}}$
• B. $\frac{A}{(x^2+a^2{)}^{3/2}}$
• C. $A(x^2+a^2{)}^{1/2}$
• D. $A(x^2+a^2{)}^{3/2}$

Answer 1

The correct option is A $\frac{A}{(x^2+a^2{)}^{1/2}}$

Given,

$E_G=\frac{Ax}{(x^2+a^2{)}^{3/2}};V_{\infty }=0$

We know that,

$E_G=-\frac{dV}{dx}$

Using above equations,

${\int }_{V_{\infty }}^{V_x}dV=-{\int }_{\infty }^x{E}_G\cdot dx$

$\Rightarrow V_x-V_{\infty }=-{\int }_{\infty }^x\frac{Ax}{(x^2+a^2{)}^{3/2}}dx$

Putting, $x^2+a^2=z$

$2xdx=dz$

$\Rightarrow V_x-0=-{\int }_{\infty }^{x^2+a^2}\frac{Adz}{(z{)}^{3/2}}$

$\Rightarrow V_x=-\frac{A}{2}{\left[\frac{z^{-\frac{3}{2}+1}}{-\frac{3}{2}+1}\right]}_{\infty }^{x^2+a^2}$

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

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(A\right)$ is the correct answer.