# Coulomb's Law

## Key Questions

• $\setminus {\vec{F}}_{12} = \frac{1}{4 \setminus \pi \setminus {\epsilon}_{o}} \frac{{q}_{1} {q}_{2}}{|} \setminus {\vec{r}}_{1} - \setminus {\vec{r}}_{2} {|}^{2} = - \setminus {\vec{F}}_{21}$

I shall outline a better method. Not just by assuming the surface is spherical with area $4 \pi {r}^{2}$.

So, to start with I shall prove it for the field due to a single point charge. (At O)
Since superposition principle applies to electric fields, this can be generalised to any number of charges quite easily.

There it is, given below.

#### Explanation:

Consider a charged particle $q$ placed at O and is being surrounded and enclosed by a closed surface $S$. (Of arbitrary size and shape).

Take an arbitrary surface element $\mathrm{dS}$ on the surface at a distance $r$ from O.

Let the normal on $\mathrm{dS}$ make an angle $\theta$ with the electric field of $q$ passing through $\mathrm{dS}$.

Thus, $\vec{E} \cdot \mathrm{dv} e c S = E \mathrm{dS} C o s \theta$ which is equal to the flux through the small area.

Now, by Coulomb's law,

$E = \frac{q}{4 \pi {\epsilon}_{0} {r}^{2}}$

Which gives, $\mathrm{dp} h i = \frac{q}{4 \pi {\epsilon}_{0}} \cdot \mathrm{dS} \cos \frac{\theta}{{r}^{2}}$

Now, $\mathrm{dS} C o s \frac{\theta}{{r}^{2}}$ is the solid angle which can be denoted as, $\mathrm{do} m e g a$

Thus, the flux through $\mathrm{dS}$ is,

$\mathrm{dp} h i = \frac{q}{4 \pi {\epsilon}_{0}} \cdot \mathrm{do} m e g a$

Integrating over the entire surface,

$\int {\int}_{S} \vec{E} \cdot \mathrm{dv} e c S = \frac{q}{4 \pi {\epsilon}_{0}} \int \mathrm{do} m e g a$

But, the net solid angle subtended by a closed surface at an internal point is always $4 \pi$.

$\int {\int}_{S} \vec{E} \cdot \mathrm{dv} e c S = \frac{q}{4 \pi {\epsilon}_{0}} \cdot 4 \pi$

Thus,
$\int {\int}_{S} \vec{E} \cdot \mathrm{dv} e c S = \frac{q}{\epsilon} _ 0$

Which is Gauss law in integral form.

One may derive the differential form from Coulomb's law as well by taking the Divergence of the electric field and then proceeding with vector calculus methods.

The differential form looks something like this, $\nabla \cdot \vec{E} = \frac{\rho}{\epsilon} _ 0$ where $\rho$ is the charge density.

$F = k \frac{{Q}_{1} , {Q}_{2}}{r} ^ 2$

#### Explanation:

assume there are two charges ${Q}_{1} , {Q}_{2}$

the force between them can be found through this relation

$F = k \frac{{Q}_{1} \cdot {Q}_{2}}{r} ^ 2$

Where k is Coulomb's constant
and $r$ is the distance between the two charges

and the force is attraction if the two charges have different sign
($+ v e \mathmr{and} - v e$)
and repulsion if the two charges have the same sign