Coulomb's law is simply a law relating the forces experienced between two point charged particles placed at a definite distance. The formal law is as follows,
Mathematically,
$$F_{E}=\frac{1}{4\pi \epsilon_{0}}\frac{Q_{1}Q_{2}}{r^{2}}$$
Where, $F_{E}=$ Electric force between the two charge, measured in Newton($N$)
$Q_{1},Q_{2}=$ Magnitude of Charge on the two particles, measured in coulomb($C$)
$\epsilon_{0}=$ Permitivity of free space.
$r=$ distance between the point particles, measured in meter($m$)
Now that we have the force law of the two charged particle one can ask what about the Electric field strength at a distance $'r'$ of the source charge. Let us consider the following figure,
The $\mathrm{Electric\ field\ Strength}$ of a point charge $Q^{+}$ at a distance $'r'$ in free space is given by,
$$E=\frac{F_{q^{+}Q^{+}}}{q^{+}}=\frac{1}{4\pi \epsilon_{0}}\frac{Q^{+}}{r^{2}}$$
where,$F_{q^{+}Q^{+}}$ is the force acting on charge $q^{+}$ due to charge $Q^{+}$ and is equal to $F_{q^{+}Q^{+}}=\frac{1}{4\pi \epsilon_{0}}\frac{q^{+}Q^{+}}{r^{2}}$.
$\mathrm{Coulomb's\ law:}$ "The magnitude of the electrostatic force of interaction between two point charges is directly proportional to the scalar multiplication of the magnitudes of charges and inversely proportional to the square of the distance between them. The force is always along the straight line joining them."-wikipedia
$$F_{E}=\frac{1}{4\pi \epsilon_{0}}\frac{Q_{1}Q_{2}}{r^{2}}$$
Where, $F_{E}=$ Electric force between the two charge, measured in Newton($N$)
$Q_{1},Q_{2}=$ Magnitude of Charge on the two particles, measured in coulomb($C$)
$\epsilon_{0}=$ Permitivity of free space.
$r=$ distance between the point particles, measured in meter($m$)
Direction of the forces: The only one thing to keep in mind is that like charges repel and the unlike charges attract,
Now that we have the force law of the two charged particle one can ask what about the Electric field strength at a distance $'r'$ of the source charge. Let us consider the following figure,
The $\mathrm{Electric\ field\ Strength}$ of a point charge $Q^{+}$ at a distance $'r'$ in free space is given by,
$$E=\frac{F_{q^{+}Q^{+}}}{q^{+}}=\frac{1}{4\pi \epsilon_{0}}\frac{Q^{+}}{r^{2}}$$
where,$F_{q^{+}Q^{+}}$ is the force acting on charge $q^{+}$ due to charge $Q^{+}$ and is equal to $F_{q^{+}Q^{+}}=\frac{1}{4\pi \epsilon_{0}}\frac{q^{+}Q^{+}}{r^{2}}$.
Here's a video on Coulomb's law. SIMPLE CONCISE AND INTUITIVE.
Respect to Prof. Eric Rogers
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