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{Ax}{{\left(x^2+a^2\right)}^{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}$ 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. $A{\left(x^2+a^2\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$
• B. $\frac{A}{{\left(x^2+a^2\right)}^{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}$
• C. $A{\left(x^2+a^2\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$
• D. $\frac{A}{{\left(x^2+a^2\right)}^{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}$

Answer 1

The correct option is B

$\frac{A}{{\left(x^2+a^2\right)}^{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}$

Step 1. Given data:

Gravitational field due to a mass distribution $\left(E_G\right)$ = $\frac{Ax}{{\left(x^2+a^2\right)}^{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}$

Step 2. Explanation:

Gravitational field due to mass distribution $E_g=\frac{Ax}{{\left(x^2+a^2\right)}^{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}$

Also gravitational potential at infinity, $V_{\infty }=0$

We have to find the value of the gravitational potential at x, $V_x=?$

As seen from the above equation, the gravitational field is varying with $x$

Let a small element at a distance $dx$ that has the potential of $dv$,

So we can write the potential value for this element.

Now Potential for this small element can be written as $dV=-E_G.dx$

Integrating both sides within limits, we get

${\int }_{\infty }^{x}dV=-{\int }_{\infty }^x{E}_G.dx$

${{\left[V\right]}^x}_{\infty }=-{\int }_{\infty }^x\frac{Ax}{{\left(x^2+a^2\right)}^{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}dx$

$V_x-V_{\infty }=-{\int }_{\infty }^x\frac{Ax}{{\left(x^2+a^2\right)}^{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}.dx$

$V_x-0=-{\int }_{\infty }^x\frac{Ax}{{\left(x^2+a^2\right)}^{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}.dx$

$V_x=-A{\int }_{\infty }^x\frac{x}{{\left(x^2+a^2\right)}^{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}.dx$

Step 3. Substituting $x$

Let $x^2+a^2=t,then2x.dx=dt$

$x.dx=\frac{dt}{2}$

At $$\begin{gathered} x=\infty ,t=\infty \\ x=x,t=x^2+a^2 \end{gathered}$$

$V_x=-A{\int }_{\infty }^{x^2+a^2}\frac{{\displaystyle \raisebox{1ex}{$dt$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}{t^{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}$

$V_x=-\frac{A}{2}{\int }_{\infty }^{x^2+a^2}{\left(t\right)}^{\raisebox{1ex}{$-3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}.dt$

$V_x=-\frac{A}{2}{{\left[\frac{{\left(t\right)}^{{\displaystyle \raisebox{1ex}{$-3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}+1}}{-{\displaystyle \frac{3}{2}}+1}\right]}_{\infty }}^{x^2+a^2}=-\frac{A}{2}{{\left[\frac{t^{{\displaystyle \raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}{{\displaystyle \frac{-1}{2}}}\right]}_{\infty }}^{x^2+a^2}=A{{\left(\frac{1}{t^{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}\right)}_{\infty }}^{x^2+a^2}$

$V_x=A\left(\frac{1}{{\left(x^2+a^2\right)}^{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}-\frac{1}{\infty }\right)=\frac{A}{{\left(x^2+a^2\right)}^{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}$

Thus, option B is the correct option.