Similar to the gravitational force between two objects you'll find Coulomb's law easier to define and express the force between two charges. We could discuss the electric field which generates the force between two charges either attractive or repulsive and then Coulomb's law but we first define Coulomb's law considering that there is a net electric field associated with an electric charge.

## Coulomb's Law

Here we denote a net point charge by symbol $q$. We consider something called *gravitational pull* which tells us that all objects on the Earth's surface are being pulled by the force of gravity. It means that there is some kind of force pulling all objects towards the centre of the Earth and we call such a force per unit mass of a body as the *gravitational field*. The gravitational field of a body is the gravitational force it exerts per unit mass of other bodies inside the influence of its field. Just like the Earth has a gravitational field which has an influence to all other objects inside its field, a fixed charge also has an *electric field* around it which also has an influence to other charges if present inside its electric field.

And similar to the gravitational field electric field of a point charge is the electric force it exerts per unit charge. You should be clear that every charge produces an electric field around it which exerts a force on other charges inside its electric field. Now consider two fixed charges ${{q}_{1}}$ and ${{q}_{2}}$ separated by a distance $r$ not very far from each other. The electric field of charge $q_1$ exerts force on charge $q_2$ and the electric field of charge $q_2$ exerts force on charge ${{q}_{1}}$. According to Newton's third law the force of charge ${{q}_{1}}$ on ${{q}_{2}}$ ($\vec F_{\text{1 on 2}}$) is exactly equal and opposite to the force of $q_2$ on $q_1$ ($\vec F_{\text{2 on 1}}$). This is the mutual interaction of charges (see Figure 1).

Now what do we know about the force between the two fixed charges? We have Coulomb's law to get the idea of the force between two fixed charges which tells us that the force between any two fixed charges $q_1$ and $q_2$ is directly proportional to the product of $q_1$ and $q_2$ and inversely proportional to the square of distance between them. That means $F \propto {q_1}{q_2}$ and $F \propto \frac{1}{{{r^2}}}$. Therefore, according to Coulomb's law

\[F = k\frac{{{q_1}{q_2}}}{{{r^2}}} \tag{1} \label{1}\]

where $k$ is the proportionality constant and $k=1/4\pi\epsilon_0$. Here $\epsilon_0$ is another constant and its value is $\epsilon_0 = 8.85\times{10^{ - 12}}\frac{{{{\rm{C}}^{\rm{2}}}}}{{{\rm{N}}{{\rm{m}}^{\rm{2}}}}}$ in three significant figures. The value of $k$ is approximately $k = 9.0 \times {10^9}\frac{{{\rm{N}}{{\rm{m}}^{\rm{2}}}}}{{{{\rm{C}}^{\rm{2}}}}}$ but its value is $k = 8.987551787 \times {10^9}\frac{{{\rm{N}}{{\rm{m}}^{\rm{2}}}}}{{{{\rm{C}}^{\rm{2}}}}}$ in more significant figures. We use the approximate value in calculations (that is, $k=9.0\times {{10}^{9}}\frac{\text{N}{{\text{m}}^{\text{2}}}}{{{\text{C}}^{\text{2}}}}$).

## Electric Field

Electric field is the electric force per unit charge. So the electric field produced by a fixed point charge $q$ at a point $p$ is the electric force per unit charge at that point $p$. Now just consider that there is a positive point charge $q_0$ at $p$ at a distance $r$ form charge $q$. The electric force between the charges $q$ and $q_0$ is

\[F = k\frac{{q{q_0}}}{{{r^2}}}\]

The electric field $E$ at the point $p$ is the electric force given by above equation divided by the point charge $q_0$.

\[E = k\frac{{q{q_0}}}{{{r^2}{q_0}}} = k\frac{q}{{{r^2}}} \tag{2} \label{2}\]

From the expression of the electric field you may notice that the electric field is independent of the charge $q_0$. So, the electric field at the point $p$ is always the same for any charge at that point. Electric field is a vector quantity and in vector form it can be expressed as,

\[\vec E = k\frac{q}{{{r^2}}}\hat r \tag{3} \label{3}\]

where $\hat{r}$ is a unit vector along the line joining the charge and point $p$. The unit vectors specify the direction in space- they point in a particular direction in space. The direction of electric field of a positive point charge is radially outward (or in the direction away from the charge) as shown in Figure 2(a) or Figure 3(a).

And the direction of electric field of a negative point charge is radially inward in the direction towards the charge as shown in Figure 2(b) or Figure 3(b). The direction of unit vector is in the same direction of electric filed of positive point charge but the direction of unit vector is opposite to the direction of electric field of negative point charge. The electric field can also be expressed in terms of electric field lines.

### Superposition of Electric Fields

We take a note on what happens to the electric field at a point when two or more point charges are present. Consider there are charges $q_1$, $q_2$, $q_3$ and $q_4$ distributed in a space and we need to find the electric field at a point $p$ due to this charge distribution (see Figure 4).

The electric field at the point $p$ is calculated by calculating the electric fields at that point by individual charges and the vector sum of those electric fields gives the net electric field (electric field due to the charge distribution) at that point. Note that you know from Eq. \eqref{2} that the electric field due to a point charge at the point $p$ is independent of any charge present at that point. The total electric field at the point $p$ due to the charge distribution is the vector sum of the electric fields due to individual charges. The electric field due to the charge $q_1$ is $\vec E_1$, the electric field due to the charge $q_2$ is $\vec E_2$ and so on. Therefore, the net electric field at the point $p$ is

\[\vec E = {{\vec E}_1} + {{\vec E}_2} + {{\vec E}_3} + {{\vec E}_4} \tag{4} \label{4}\]

Also note that $\vec F = q_0\vec E$ and if a charge $q_0$ is present at the point $p$, the total electric force at that point due to the charge distribution is

\[\begin{align*} \vec F &= {q_0}\vec E = {q_0}{{\vec E}_1} + {q_0}{{\vec E}_2} + {q_0}{{\vec E}_3} + {q_0}{{\vec E}_4}\\ \therefore \quad \vec F &= {{\vec F}_1} + {{\vec F}_2} + {{\vec F}_3} + {{\vec F}_4} \tag{5} \label{5} \end{align*}\]

Hence the total electric force at the point $p$ on charge $q_0$ is the vector sum of the electric forces due to individual charges which should be.