Application of Gauss Law

Application of Gauss Law:

1. Gravitational field intensity due to a spherical shell:

Consider a spherical shell of mass M and radius R. Let us take a point P at a distance r from the centre of the shell.

Case(a): When the point P lies outside the surface of the shell:

Let E be the gravitational field intensity at point P due to shell.

Let us draw a sphere of radius r with the centre at the centre of the shell of radius R such that the gravitational field intensity at any pointr on its surface is the same. Then by the definition flux

\(\Phi =\vec{E}.\vec{A}\)\begin{align*}=E.4\pi r^{2}\rightarrow eqnI \end{align*}Again, Applying Gauss law,\begin{align*}\Phi=-4\pi G.mass\space enclosed\end{align*}\begin{align*}=-4\pi GM\rightarrow eqnII \end{align*}from(I) and (II), we get\begin{align*}E.4\pi r^{2}=-4\pi GM \end{align*}\begin{align*}E=-\frac{GM}{r^{2}} \end{align*}

Case(b): When the point P lies on the surface of the shell:

We know, field intensity when P lies outside the shell is\begin{align*}E=-\frac{GM}{r^{2}} \end{align*}But on the surface r=R,\begin{align*}\therefore E=-\frac{GM}{R^{2}} \end{align*}

Case (c): When the point P lies inside the shell

For this construct Gaussian surface of radius r as shown in figure. Then, by definition,\begin{align*}Gravitational\space flux(\Phi)= \vec{E}.\vec{A}\end{align*}\begin{align*}=E.4\pi r^{2}\rightarrow eqnIII \end{align*}Using Gauss law, gravitational flux flowing out of the surface of radius r is,\begin{align*}\Phi =-4\pi G.mass\space enclosed\space by\space the\space surface\space of\space radius\space r \end{align*}\begin{align*} =-4\pi G.0\end{align*}\begin{align*}=0\rightarrow eqnIV \end{align*}from(III) and (IV)\begin{align*}E=0 \end{align*}

2. Gravitational field intensity due to a solid sphere:

Consider a solid sphere of mass M and radius R. Let us take a point P where we are going to find the field intensity due to sphere.

Case(I): when the point P lies outside the sphere.

case-I

Let the point P is at a distance r from the centre of the sphere as shown in figure. Let us draw a spherical surface of radius r with the centre at the centre of the sphere.

By definition,

Gravitational flux

\begin{align*}\Phi =\vec{E}.\vec{A}=E.4\pi r^{2}\rightarrow eqn(I) \end{align*}where E is the field intensity at point P.

Also from Gauss Law\begin{align*}\Phi=-4\pi G.mass\space enclosed\rightarrow eqn(II) \end{align*}Equating (I) and (II),\begin{align*}E.4\pi r^{2}=-4\pi GM \end{align*}\begin{align*}E=-\frac{GM}{r^{2}} \end{align*}

Case(II): When the point P lies on the surface of the sphere, r=R

\begin{align*}\therefore Field\space intensity\space at\space point\space P\space on\space the\space surface=-\frac{GM}{R^{2}}\end{align*}

Case(III): When the point P lies inside the sphere:

Let the point P lie at a distance r from the centre of the sphere inside the sphere. Let us draw a Gaussian surface of radius r with the centre at the centre of the sphere.

Then from Gauss law, the gravitational flux over the surface of radius r is \begin{align*}\Phi =-4\pi G.mass\space enclosed \end{align*}\begin{align*}=-4\pi G.\frac{4}{3}\pi r^{3}\rho \end{align*}where \(\rho \) is the density of material of the sphere.

By definition, Gravitational flux,\begin{align*}\Phi=E.4\pi r^{2} \end{align*}\begin{align*} \end{align*}E=field intensity at point P

Equating (I) and (II) we get\begin{align*}E.4\pi r^{2}=-4\pi G.\frac{4}{3}\pi r^{3}\rho \end{align*}\begin{align*}E=-\frac{GMr}{R^{3}} \end{align*}where M=\(\frac{4}{3}\pi r^{3}\rho \)

References

Adhikari, Pitri Bhakta. A Textbook of Physics Volume-I. Kathmandu: Sukunda Pustak Bhawan, 2015.

Feynman, Richard P. The Feynman Lectures on Physics Volume 1. Noida: Dorling Kindersley (India) Pvt. Ltd., 2014.

Mathur, D S. Mechanics. New Delhi: S. Chand & Company Pvt. Ltd., 2015.

Young, Hugh D, Roger A Freedman and A Lewis Ford. University Physics. Noida: Dorling Kindersley (India) Pvt. Ltd., 2014.