Question

The gravitational field due to a disc of mass $M$ and radius $R$ at a point located $x$ distance away from the centre of disc along the axis of disc is given by

• A. $\frac{2GMx}{R^2}[\frac{1}{x}-\frac{1}{\sqrt{R^2+x^2}}]$
• B. $\frac{2GMx}{R^2}\left[\frac{1}{\sqrt{R^2+x^2}}\right]$
• C. $\frac{GMx}{R^2}\left[\frac{1}{\sqrt{R^2+x^2}}\right]$
• D. $\frac{GMx}{R^2}[\frac{1}{x}-\frac{1}{\sqrt{R^2+x^2}}]$

Answer 1

The correct option is A $\frac{2GMx}{R^2}[\frac{1}{x}-\frac{1}{\sqrt{R^2+x^2}}]$

Let us choose an elementary ring of thickness $dr$ at a distance $r$ from the centre.


So, the mass of the element,

$dm=\frac{M}{\pi R^2}\left(2\pi rdr\right)=\frac{2Mrdr}{R^2}$

The field at any axial point at distance $x$ due to the elementry disc is given by,

$dE=\frac{G\left(dm\right)x}{(r^2+x^2{)}^{3/2}}$

The gravitational field due to the whole disc will be,

$E_T=\int dE$

$\Rightarrow E_T={\int }_{x=0}^{x=R}\frac{G\left(dm\right)x}{(r^2+x^2{)}^{3/2}}$

Substituting the value of $dm$,

$\Rightarrow E_T={\int }_{x=0}^{x=R}\frac{2GxM}{R^2}\frac{rdr}{(r^2+x^2{)}^{3/2}}$

Substituting,
$\left[\right(r^2+x^2)=t^2\Rightarrow rdr=tdt]$

$\Rightarrow E_T=\frac{2GxM}{R^2}{\int }_{t=x}^{t=\sqrt{R^2+x^2}}\frac{tdt}{(t^2{)}^{3/2}}$

$\Rightarrow E_T=\frac{2GxM}{R^2}{\int }_{t=x}^{t=\sqrt{R^2+x^2}}\frac{dt}{t^2}$

$\Rightarrow E_T=\frac{2GxM}{R^2}{[-\frac{1}{t}]}_x^{\sqrt{R^2+x^2}}$

$\Rightarrow E_T=\frac{2GxM}{R^2}[\frac{1}{x}-\frac{1}{\sqrt{R^2+x^2}}]$

Hence, option (a) is correct answer.