Coulomb's Law & Superposition Principle

Coulomb’s Law: The Push and Pull of Charges

We know that like charges repel and unlike charges attract, but how much exactly? In 1785, Charles Augustin de Coulomb used a torsion balance to measure this force.

The Law States: The electrostatic force of attraction or repulsion between two stationary point charges is directly proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance between them. This force acts along the line joining the two charges.

Mathematical Expression

If two point charges $q_1$ and $q_2$ are separated by a distance $r$ in vacuum, the magnitude of the force $F$ is: $$F = k \frac{|q_1 q_2|}{r^2}$$

Where $k$ is the electrostatic force constant. Its value depends on the medium between the charges.

For Vacuum/Air: $$k = \frac{1}{4\pi\epsilon_0} \approx 9 \times 10^9 \text{ Nm}^2/\text{C}^2$$

Here, $\epsilon_0$ (Epsilon-nought) is the permittivity of free space. $$\epsilon_0 = 8.854 \times 10^{-12} \text{ C}^2/\text{Nm}^2$$

Direction of Coulomb Force

Force is a vector quantity, so direction matters.

Case 1: Like Charges

If both charges have the same sign:

• positive-positive or negative-negative,
• the force is repulsive.

Case 2: Unlike Charges

If the charges have opposite signs:

• positive-negative,
• the force is attractive.

Important Direction Rule

The force always acts along the line joining the two charges.

This makes Coulomb force a central force.

Scalar Form of Coulomb’s Law

When only magnitude is required, use: $$F = \frac{1}{4\pi \varepsilon_0} \frac{|q_1 q_2|}{r^2}$$

This formula gives only the magnitude of force.

To identify attraction or repulsion, use the sign of charges separately.

Vector Form of Coulomb’s Law

Force is a vector, so direction matters!

Suppose charges $q_1$ and $q_2$ are located at position vectors $\vec{r}_1$ and $\vec{r}_2$ respectively.

Then the vector from charge 1 to charge 2 is: $$\vec{r}_{21} = \vec{r}_2 - \vec{r}_1$$ Its magnitude is $r_{21}$ and the corresponding unit vector is: $$\hat{r}_{21} = \frac{\vec{r}_{21}}{r_{21}}$$

Then the force on charge 2 due to charge 1 is: $$\vec{F}_{21} = \frac{1}{4\pi \varepsilon_0} \frac{q_1 q_2}{r_{21}^2} \hat{r}_{21}$$

Similarly, the force on charge 1 due to charge 2 is: $$\vec{F}_{12} = \frac{1}{4\pi \varepsilon_0} \frac{q_1 q_2}{r_{12}^2} \hat{r}_{12}$$

Since: $$\hat{r}_{12} = -\hat{r}_{21}$$ we get: $$\vec{F}_{12} = -\vec{F}_{21}$$

This proves that electrostatic forces obey Newton’s Third Law of Motion.

Meaning of the Sign in Vector Form

In vector form, we do not need to write separate formulas for attraction and repulsion.

That is because the product $q_1 q_2$ automatically determines the direction.

If $q_1 q_2 > 0$

Then force is along $\hat{r}_{21}$. This corresponds to repulsion.

If $q_1 q_2 < 0$

Then force is opposite to $\hat{r}_{21}$. This corresponds to attraction.

Effect of Medium (Relative Permittivity)

If the charges are placed in a medium (like water or oil), the force decreases. We define Relative Permittivity ($K$ or $\epsilon_r$), also known as the Dielectric Constant: $$K = \epsilon_r = \frac{\epsilon}{\epsilon_0} = \frac{F_{\text{vacuum}}}{F_{\text{medium}}}$$

For example, if you put charges in water ($K \approx 80$), the force becomes $1/80$th of what it was in the air!

Principle of Superposition

What if there are more than two charges? Coulomb's law only handles pairs. To find the net force on a specific charge due to multiple other charges, we use the Principle of Superposition.

It states: The total force on any charge due to a number of other charges is the vector sum of all the forces exerted on it by those charges, taken one at a time. The individual forces are unaffected by the presence of other charges.

$$\vec{F}_{total} = \vec{F}_1 + \vec{F}_2 + \vec{F}_3 + \dots + \vec{F}_n$$

Problem-Solving Strategy

Step 1

Write the known values of $q_1$, $q_2$, and $r$ in SI units.

Step 2

Use the scalar form for magnitude: $$F = 9 \times 10^9 \frac{|q_1 q_2|}{r^2}$$

Step 3

Decide nature:

• same sign: repulsion,
• opposite sign: attraction.

Step 4

If the problem is one-dimensional, assign direction using sign convention.

Step 5

If vector form is needed, write force along unit vector.

🧠 Memory Capsule

• Inverse Square Law: If distance $r$ doubles, Force becomes $1/4$th. If distance is halved, Force becomes 4 times.
• Point Charges: Coulomb's Law is strictly valid for point charges at rest.
• Direction: Force is always along the line joining the centers of the two charges.
• Medium: $F_{\text{medium}} = \frac{F_{\text{air}}}{K}$. Force is maximum in vacuum.
• Superposition: Treat each force as a separate vector, then use the parallelogram law or component method to add them.

Solved Examples

Example 1: Basic Calculation

Calculate the electrostatic force between two protons separated by a distance of $1.6 \times 10^{-15}$ m (typical nuclear distance).

Solution:

1. Identify values: $q_1 = q_2 = 1.6 \times 10^{-19}$ C. $r = 1.6 \times 10^{-15}$ m. $k = 9 \times 10^9$.
2. Apply Formula: $F = k \frac{q_1 q_2}{r^2}$.
3. Substitute: $F = (9 \times 10^9) \frac{(1.6 \times 10^{-19})^2}{(1.6 \times 10^{-15})^2}$.
4. Simplify: $F = (9 \times 10^9) \frac{2.56 \times 10^{-38}}{2.56 \times 10^{-30}} = 9 \times 10^9 \times 10^{-8} = 90$ N.
5. Result: The force is 90 N (Repulsive).

Example 2: Change in Distance

Two charges attract each other with a force of 100 N. If the distance between them is tripled, what will be the new force?

Solution:

1. Relationship: $F \propto 1/r^2$.
2. Ratio Method: $\frac{F_2}{F_1} = (\frac{r_1}{r_2})^2$.
3. Substitute: Since $r_2 = 3r_1$, the ratio is $(1/3)^2 = 1/9$.
4. Calculate: $F_2 = 100 / 9 \approx 11.11$ N.
5. Result: The force reduces to 11.11 N.

Example 3: Effect of Dielectric

The force between two charges in air is 80 N. When a dielectric slab is placed between them, the force drops to 10 N. What is the dielectric constant of the slab?

Solution:

1. Formula: $K = F_{\text{air}} / F_{\text{medium}}$.
2. Substitute: $K = 80 / 10 = 8$.
3. Result: The dielectric constant is 8.

Example 4: Null Point (JEE/NEET Level)

Two point charges $+9e$ and $+e$ are kept 16 cm apart. At what distance from charge $+e$ should a third charge $q$ be placed so that it stays in equilibrium?

Solution:

1. Logic: For equilibrium, forces from both charges must be equal and opposite. The third charge must be between them on the line joining them.
2. Setup: Let $x$ be the distance from $+e$. Then distance from $+9e$ is $(16 - x)$.
3. Equation: $k \frac{(9e)q}{(16-x)^2} = k \frac{(e)q}{x^2}$.
4. Simplify: $\frac{9}{(16-x)^2} = \frac{1}{x^2} \Rightarrow \frac{3}{16-x} = \frac{1}{x}$ (Taking square root).
5. Solve: $3x = 16 - x \Rightarrow 4x = 16 \Rightarrow x = 4$ cm.
6. Result: 4 cm from the $+e$ charge.

Example 5: Superposition on a Triangle

Three equal charges $q$ are placed at the corners of an equilateral triangle of side $a$. Find the net force on any one charge.

Solution:

1. Identify forces: On one charge, two other charges exert forces $F_1$ and $F_2$.
2. Magnitude: $F_1 = F_2 = k \frac{q^2}{a^2} = F$.
3. Angle: The angle between $F_1$ and $F_2$ is $60^\circ$ (equilateral triangle).
4. Vector Addition: $F_{\text{net}} = \sqrt{F^2 + F^2 + 2F^2 \cos 60^\circ} = \sqrt{F^2 + F^2 + F^2} = \sqrt{3}F$.
5. Result: $F_{\text{net}} = \sqrt{3} \frac{kq^2}{a^2}$ along the bisector of the angle.

Example 6: Comparing Forces

Compare the electrostatic force and gravitational force between two electrons kept at a distance $r$.

Solution:

1. Formulas: $F_e = k \frac{e^2}{r^2}$ and $F_g = G \frac{m_e^2}{r^2}$.
2. Ratio: $\frac{F_e}{F_g} = \frac{k e^2}{G m_e^2}$.
3. Substitute: $\frac{(9 \times 10^9) (1.6 \times 10^{-19})^2}{(6.67 \times 10^{-11}) (9.1 \times 10^{-31})^2}$.
4. Calculate: The ratio is approximately $10^{42}$.
5. Result: Electrostatic force is immensely stronger than gravitational force at the atomic level!

Example 7: Minimum Force Possible

What is the minimum electrostatic force between two charged bodies kept at a distance of 1 m?

Solution:

1. Logic: Force is minimum when charges are minimum. The minimum possible charge is $e$.
2. Calculation: $F = k \frac{e^2}{r^2} = (9 \times 10^9) \frac{(1.6 \times 10^{-19})^2}{1^2}$.
3. Result: $F = 2.304 \times 10^{-28}$ N.

Example 8: System Equilibrium

Two identical spheres having charges $q$ are suspended by strings of length $L$. If the angle of divergence is $2\theta$, find $q$ in terms of tension $T$.

Solution:

1. Analyze forces: Gravity ($mg$) down, Tension ($T$) along string, Electrostatic force ($F_e$) horizontal.
2. Components: $T \cos \theta = mg$ and $T \sin \theta = F_e$.
3. Relate: $F_e = T \sin \theta$.
4. Substitute Force: $k \frac{q^2}{r^2} = T \sin \theta$.
5. Result: $q = \sqrt{\frac{r^2 T \sin \theta}{k}}$.

Example 9: Superposition on a Square

Four charges $+q, +q, -q, -q$ are placed at the corners of a square of side $L$. Find the force on a charge $Q$ placed at the center.

Solution:

1. Geometry: Center is distance $r = L/\sqrt{2}$ from each corner.
2. Symmetry: Forces from the two $+q$ charges act away from the corners. Forces from the two $-q$ charges act toward the corners.
3. Pairing: The two $+q$ charges and two $-q$ charges are opposite each other. Forces from like pairs ($+q$ and $+q$) cancel out if $Q$ is central and all charges are same. But here we have $+q, +q, -q, -q$.
4. Vector Sum: The forces from the two $+q$ charges point toward the $-q$ charges. Sum the vectors carefully based on the specific arrangement of the corners.
5. Result: Usually results in $4\sqrt{2} \frac{kqQ}{L^2}$ depending on charge positions.

Example 10: Percentage Change

If the charge on both bodies is doubled and the distance between them is doubled, what is the percentage change in the force?

Solution:

1. Initial: $F_1 = k \frac{q_1 q_2}{r^2}$.
2. Final: $F_2 = k \frac{(2q_1)(2q_2)}{(2r)^2} = k \frac{4 q_1 q_2}{4 r^2} = F_1$.
3. Result: $F_2 = F_1$. There is 0% change.