Coulomb's Law

Coulomb’s law helps us to determine the magnitude and direction of an electrostatic force between two point charges. We need to keep in mind that it works only on point charges.

Definition of Coulomb’s Law

The magnitude of electrostatic force of attraction or repulsion between two point charges is directly proportional to the product of magnitudes of two charges and is inversely proportional to the square of distance between their centers. The force is directed along the line joining the centers of the two charges.

For two charges $Q_1$ and $Q_2$ at distance $r$
magnitude of electrostatic force will be
$F \propto Q_1 Q_2$ and $F \propto \frac{1}{r^2}$
or
$F \propto \frac{Q_1 Q_2}{r^2}$
or $F = k \frac{Q_1 Q_2}{r^2}$
where $k$ is a proportionality constant whose value depends upon the nature of medium between charges.
value of $k$ for air or vacuum is $9 \times 10^9 \, \text{N m}^2 \text{C}^{-2}$.

Vector Form of Coulomb’s Law

• On both charges, forces are equal and opposite (like action and reaction).
• Nature of charge (positive or negative) is used to find the direction of force only. Its sign is not used while calculating magnitude of force. (or in the formula of F, the sign of q is not used).

Dependence of Electrostatic Force on Surrounding Medium

Electrostatic force between two charge is given by
$F = k \frac{Q_1 Q_2}{r^2}$

$F$ depends on constant $k$,
$k$ depends on the surrounding medium to the charge.
For simplicity we write: $k = \frac{1}{4\pi\epsilon_0}$

$\epsilon_0$ is read as Epsilon Zero.
where $\epsilon_0$ is known as Absolute permittivity of free space (air or vacuum).

for any other medium: $k = \frac{1}{4\pi\epsilon}$
where $\epsilon$ is known as Absolute permittivity of the medium.

Absolute permittivity is the property of a medium and
now force between two charges kept in a medium of permittivity $\epsilon$
$F = \frac{1}{4\pi\epsilon} \frac{Q_1 Q_2}{r^2}$
clearly, $F \propto \frac{1}{\epsilon}$.
Therefore, higher the value of $\epsilon$, less the electrostatic force between charges.

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⤵️⤵️⤵️

Let two charges $Q_1$ and $Q_2$ are kept in air at distance $r$.
Electrostatic force between them ⟶
$F_0 = \frac{1}{4\pi\epsilon_0} \frac{Q_1 Q_2}{r^2}$

When the same two charges are kept at the same distance in a medium of absolute permittivity $\epsilon$.

Electrostatic force will be ⟶
$F(m) = \frac{1}{4\pi\epsilon} \frac{Q_1 Q_2}{r^2}$

Now dividing $F(m)$ by $F_0$
$\frac{F(m)}{F_0} = \frac{\epsilon_0}{\epsilon}$
here $\frac{\epsilon_0}{\epsilon}$ is a constant and we write $\frac{\epsilon_0}{\epsilon} = \frac{1}{K}$ or $K = \frac{\epsilon}{\epsilon_0}$, where $K$ is known as relative permittivity of the medium or dielectric constant of that medium.

Therefore, $\frac{F(m)}{F_0} = \frac{1}{K}$ or $F(m) = \frac{F_0}{K}$.

Dielectric Constant (Relative Permittivity)

Dielectric constant K is the property of a medium which affects electric field and electrostatic force. Note that dielectric constant K is different from the k used in the formula of Coulomb’s law.

Definition 1

Dielectric constant of a medium is defined as the ratio of the Absolute permittivity of the medium to that of the permittivity of air or vacuum.

$K = \frac{\epsilon}{\epsilon_0}$

Since, dielectric constant is a ratio of similar quantities, it has no unit.

For air or vacuum, $K=1$. It means $\epsilon = \epsilon_0$, only for air or vacuum.
For water, $K=81$. And this is the reason why bond strength ionic compound decreases by 81 times when it enters in water.

Absolute permittivity of any medium or $\epsilon$ is $\epsilon_0 K$.

Definition 2

Dielectric constant of a medium is defined as the ratio of electrostatic force between two point charges kept in air or vacuum to that between two point charges kept in another medium at the same distance.

$K = \frac{F_0}{F(m)}$

Absolute permittivity of any dielectric medium (insulating medium or other than conductors) is greater than that of air or vacuum.

It means, $\frac{F_0}{F(m)} > 1$
or $F_0 > F(m)$

Hence, force between two charges when kept in a dielectric medium decreases. And it decreases by $K$ times, the dielectric constant.

Dielectric Medium

• Insulating or non conducting.
• Have no free electrons.
• Mica, glass, plastic, oil, paper, etc.

Coulomb’s Law formula for a Dielectric Medium of constant K

$F(m) = \frac{1}{4\pi\epsilon_0 K} \frac{Q_1 Q_2}{r^2}, \quad \text{since } \epsilon = \epsilon_0 K$
also, $F \propto \frac{1}{K}$.

Principle Of Superposition (for Electrostatic Force)

This principle is used to determine the net force on a point charge due to multiple charges in surroundings.

👉 The net electrostatic force experienced by a point charge due to a number of charges around it is the vector sum of all the electrostatic forces acting on it due to all point charges individually.
$\vec{F}_{\text{net}} = \vec{F}_1 + \vec{F}_2 + \vec{F}_3 + \dots$ (do the vector sum).

Value, Unit and Dimensional formula for ∈₀

Value of $\epsilon_0$ is $8.85 \times 10^{-12} \, \text{N}^{-1}\text{m}^{-2}\text{C}^2$.

Dimensional formula for $\epsilon_0$ is
$[\epsilon_0] = \frac{[Q_1][Q_2]}{[4][\pi][F][r]^2}$

Since, 4 and $\pi$ are mathematical numbers, they do not have any specified dimensions. So the actual result will come from

$[\epsilon_0] = \frac{[Q_1][Q_2]}{[F][r]^2}$
$[\epsilon_0] = \frac{[AT][AT]}{[MLT^{-2}][L^2]}$
$[\epsilon_0] = \frac{[A^2T^2]}{[ML^3T^{-2}]}$
$[\epsilon_0] = [M^{-1}L^{-3}T^4A^2]$

Variations In Graph

$F = \frac{1}{4\pi\epsilon_0 K} \frac{Q_1 Q_2}{r^2}$

F vs r ($Q_1$ and $Q_2$ are constant)

$F \propto \frac{1}{r^2}$, inverse square relation.
$F$ is inversely proportional to $r$.

Graph shows “inverse square relation between F and r.”

F vs $Q_1 Q_2$ ($r$ is constant)

$F \propto Q_1 Q_2$
it is directly proportional, that is why the graph is a straight line.

Graph shows “a straight line, showing as $Q_1 Q_2$ increases or decreases $F$ also increases or decreases.”

Slope of this graph is : $\frac{F}{Q_1 Q_2} = \frac{1}{4\pi\epsilon_0 K r^2}$

F vs $1/r^2$ ($Q_1 Q_2$ is constant)

$F \propto \frac{1}{r^2}$, $F$ is directly proportional to $\frac{1}{r^2}$.

So the graph will be a straight line.

Graph shows ”$F$ is directly proportional to $\frac{1}{r^2}$, hence the graph is a straight line.”

Slope of this graph is, $\frac{F}{1/r^2} = \frac{Q_1 Q_2}{4\pi\epsilon_0 K}$

This slope depends on $Q_1 Q_2$ and here it is kept constant and therefore the whole ratio becomes a constant, this constant ratio is the slope of this graph. Since the slope is constant, the graph will be a straight line. This is also a way to find whether the graph is a straight line or not.