Question

State Gauss theorem and uses it to derive the Coulomb's inverse square law.

Answer 1

Gauss's Theorem: According to Gauss's theorem total number of electric lines of force passing normally through a closed surface of ray shape in an electric field(i.e., total electric flux) is equal to $1{/}_{{\epsilon }_o}$ times the total charge present within that surface.

i.e., ${\Phi }_E=\frac{q}{{\epsilon }_o}$

where, ${\epsilon }_o=$ permitivity if free space, q in vaccum.

Derivation of Gauss's Theorem: Let $+$q charge is placed at a point O and a point P lies at distance r from the point O. Imagine a sphere of radius r and centre O. Thus, point P lies on the surface of the sphere. Now, the surface of the sphere will be have as Gaussian surface. Therefore, the intensity of electric field on the surface at all the points will be equal in magnitude and will be directed radially outward.

$\therefore$ The electric flux passing through the spherical surface.

$\Phi =E.S.cos{0}^o$

Where S is surface area i.e., $S=4\pi r^2=$ Surface area of sphere

$\therefore {\Phi }_E=E.S.$

or ${\Phi }_E=4\pi r^{2}E$ .........(i)

But, by Gauss, theorem, we have

${\Phi }_E=\frac{q}{{\epsilon }_o}$ .....(ii)

Hence, from equation (i) and (ii), we get

$E4\pi r^2=\frac{q}{{\epsilon }_o}$

or $E=\frac{q}{4\pi {\epsilon }_o{r}^2}$

Now, imagines a charge $q_o$ placed at point P.

$\therefore$Force on $q_o$,

$F=q_{o}E$

$F=\frac{q_{o}q}{4\pi {\epsilon }_o{r}^2}$

$F=\frac{1}{4\pi {\epsilon }_o}\cdot \frac{q_{o}q}{r^2}$.

Which is Coulomb's inverse square law.