Coulomb’s Law

Coulomb’s Law states that the force of attraction or repulsion between two point charges is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.

According to Coulomb’s law, the force of attraction or repulsion between two charged bodies is directly proportional to the product of their charges and inversely proportional to the square of the distance between them. It acts along the line joining the two charges considered to be point charges.

Table of Contents

• Coulomb’s Law Formula
• Coulomb’s Law in Vector Form
• What is One Coulomb of Charge?
• Key Points on Coulomb’s Law
• Limitations of Coulomb’s Law
• Relative Permittivity of a Material
• Applications
• Problems on Coulomb’s Law

Coulomb’s Law Formula

In Short: $F\propto \frac{q_1{q}_2}{d^2}$

Where

• ε is the absolute permittivity
• K or ε_r is the Relative Permittivity or Specific Inductive Capacity.
• The permittivity of free space is represented by ε₀.
• ε_r, also known as the dielectric constant of the medium, is the ratio of the electric field in the medium to the electric field in a vacuum when two charges are placed in the medium.

History of Coulomb’s Law

In 1785, Charles Augustin de Coulomb, a French physicist, published an equation known as Coulomb’s Law or Coulomb’s Inverse-Square Law, which established a tangible mathematical relationship between two bodies that have been electrically charged, describing the force causing them to attract or repel each other.

Coulomb’s Law in Vector Form

$\begin{array}{l}{\overrightarrow{F}}_{12}=\frac{1}{4\pi {\in }_0}\frac{q_1{q}_2}{r_{12}^2}{\hat{r}}_{12};{\overrightarrow{F}}_{12}=-{\overrightarrow{F}}_{21}\end{array}$

F₁₂ is the force exerted by q₁ on q₂, and F₂₁ is the force exerted by q2 on q1.

Coulomb’s law holds for stationary charges only, which are point sized and obeys Newton’s third law.

$\begin{array}{l}(ie{\overrightarrow{F}}_{12}=-{\overrightarrow{F}}_{21})\end{array}$

The force on a charged particle due to a number of point charges is the result of forces from each individual point charge, i.e. the sum of the forces from each point charge.

$\overrightarrow{F}={\overrightarrow{F}}_1+{\overrightarrow{F}}_2+{\overrightarrow{F}}_3+\dots$

What is 1 Coulomb of Charge?

A Coulomb of charge is the amount of electric charge transported by a constant current of one ampere in one second. It is equivalent to the charge of 6.241 x 10¹⁸ electrons.

The Coulomb force is the conservative mutual and internal force that repels an equal charge of the same sign with a force of 9×10⁹ N when the charges are one meter apart in a vacuum.

The value of ε_o is 8.86 × 10^-12 C²/Nm² (or) 8.86 × 10^-12 Fm^–1

Note: Coulomb force is true only for static charges, which remain at rest.

Coulomb’s Law - Conditions for Stability

If q is slightly displaced towards A, F_A increases in magnitude while F_B decreases in magnitude. Now the net force on q is toward A so it will not return to its original position. So for axial displacement, the equilibrium is unstable.

If q is displaced perpendicularly to AB, the forces F_A and F_B bring the charge back to its original position. Therefore, for a perpendicular displacement, the equilibrium is stable.

Key Points on Coulomb’s Law

1. Coulomb’s Law states that the force of attraction or repulsion between two electric charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.
2. The SI unit of charge is the coulomb (C).
3. The magnitude of the electrostatic force between two charges is given by the equation F=kq_1q2/r², where k is Coulomb’s constant, q1 and q2 are the magnitudes of the two charges, and r is the distance between them.
4. The direction of the force is along the line joining the two charges, and is attractive if the charges are of opposite sign, and repulsive if the charges are of the same sign.

1. If the force between two charges in two different media is the same for different separations, $$F=\frac{1}{K}\frac{1}{4\pi {\epsilon }_0}\frac{q_1{q}_2}{r^2}=constant$$

2. K₁r₁² = K₂r₂²

3. From Coulomb’s Law, if the force between two charges separated by a distance r₀ in a vacuum is the same as the force between the same charges separated by a distance r in a medium, then Kr² = r₀²

4. Two identical conductors having charges q₁ and -q₂ are put to contact and then separated, after which each will have a charge equal to (q₁ - q₂)/2. If the charges are q₁ and q₂, each will have a charge equal to (q₁ + q₂)/2.

5. The charges of the two spherical conductors, having charges q₁ and q₂ and radii r₁ and r₂, after contact and then being separated are:

$q_1=[r_1/(r_1+r_2)]\text{*}(q_1+q_2)and{q}_2=[r_2/(r_1+r_2)](q_1+q_2)$

6. If the force of attraction or repulsion between two identical conductors having charges $q_1$ and $q_2$ when separated by a distance $d$ is $F$, then if they are put in contact and then separated by the same distance, the new force between them is: $$F = \frac{F\left( q_1+q_2 \right)^2}{4q_1q_2}$$

**7.**If charges are q₁ and -q₂, then

$F=\frac{F(q_1+q_2{)}^2}{4q_1{q}_2}$

8. The ratio of the electrical force to the gravitational force between two electrons separated by a certain distance is 10⁴².

9. The ratio of the electrical force to the gravitational force between two protons separated by a certain distance is 10³⁶.

10. The ratio of the electrical force to the gravitational force between a proton and an electron separated by a certain distance is ³⁹:1.

11. The expression $c=1/\surd ({\mu }_o{\epsilon }_o)$ gives the relationship between the velocity of light, the permeability of free space, and the permittivity of free space.

12. If Coulomb’s law is applied to two identical balls of mass m, suspended by silk threads of length l from the same hook and carrying similar charges q, then:

• The distance between balls $[\frac{q^{2}2l}{4\pi {ϵ}_{o}mg}{]}^{\frac{1}{3}}$

The tension in the thread $\sqrt{f^2+(mg{)}^2}$

If the total system is kept in space, then the angle between threads is 180°, and the tension in a thread is given by:

$$\begin{array}{l}T=\frac{q^2}{4\pi {ϵ}_0{\ell }^2}\end{array}$$

The electrostatic force between a charge Q divided into q and (Q - q) is maximum when

$$\begin{array}{l}\frac{q}{Q}=\frac{1}{2}\text{or}\frac{q}{Q-q}=1\end{array}$$

Limitations of Coulomb’s Law

Coulomb’s Law is limited in its ability to accurately predict the force of attraction between two objects in certain scenarios, such as when the objects are charged with different types of particles or when the objects are close together. Additionally, Coulomb’s Law does not take into account the effects of relativity.

The law only applies to point charges that are not in motion.

Coulomb’s Law can only be applied when the Inverse Square Law is followed.

It is difficult to implement Coulomb’s law when charges are in an arbitrary shape, as it is difficult to determine the distance between the charges in such cases.

The charge on the big planets cannot be calculated using the law directly.

The Dielectric Constant of a Material

$\begin{array}{l}{\epsilon }_r=K=\frac{forcebetweentwochargeinair}{forcebetweensamechargeinthemediumatthesamedistance}\end{array}$

$${ϵ}_r=\frac{F_a}{F_m}$$

For air, K = 1

**For metals, K = K = infinity

The force between 2 charges depends upon the nature of the intervening medium, whereas gravitational force is independent of the intervening medium.

For air or vacuum, $F=\frac{1}{4\pi {\epsilon }_o}.\frac{q_1}{q_2}{d}^2$

The value of $\frac{1}{4\pi {ϵ}_0}isequalto9\times {10}^9{\text{Nm}}^2/{\text{C}}^2$

⇒ Related Topics:

Electrostatics

Electric Charge

Gauss’s Law

Application of Coulomb’s Law

To determine the distance and force between two charges.

The Coulomb’s Law can be used to calculate the electric field.

$$\begin{array}{l}E=\frac{F}{Q_T}\times \left(\frac{N}{C}\right)\end{array}$$

Where E = Strength of the Electric Field

F = Electrostatic Force

Q_T = Test charge in coulombs

To calculate the force on one point due to the presence of several points using the Theorem of Superposition.

Problems on Coulomb’s Law

Problem 1: Charges of magnitude 100 microcoulombs each are located in a vacuum at the corners A, B, and C of an equilateral triangle measuring 4 meters on each side. If the charge at A and C are positive and the charge at B is negative, what is the magnitude and direction of the total force on the charge at C?

Sol.

Fig. shows the situation. Let us analyze the forces acting on C due to A and B.

The force of repulsion on C due to A, i.e., FCA in the direction of AC, can be calculated using Coulomb’s law.

$$\begin{array}{l}{F}_{AC}=\frac{1}{4\pi {\epsilon }_0}.\frac{q\times q}{a^2}\end{array}$$ along AC

The force of attraction on C due to B, i.e., FCB in direction CB, is given by this formula.

$$\begin{array}{l}{F}_{CB}=\frac{1}{4\pi {\epsilon }_0}\frac{q^2}{a^2}\end{array}$$ along the direction of the Coulombic force (CB)

Thus, the two forces are equal in magnitude and the angle between them is 120º. The resultant force F can be calculated using the following equation:

$\begin{array}{l}F=\sqrt{F_{C_A}^2+F_{C_B}^2+2F_{C_A}\times F_{C_B}\cos {120}^{\text{o}}}\end{array}$

$\frac{q^2}{4\pi \epsilon a^2}=\frac{9\times {10}^9\times {(100\times {10}^{-6})}^2}{4\pi \times 8.85\times {10}^{-12}\times {(100\times {10}^{-6})}^2}=5.625\text{Newton}$

This force is parallel to AB.

Problem 2: Determine the position and magnitude of a positive point charge q such that the system consisting of a negative point charge of unit magnitude and the positive point charge q is in equilibrium. Analyze the system to determine if it is a stable, unstable, or neutral equilibrium.

The two negative charges A and B of unit magnitude, as shown in Fig., are separated by a distance of r_A from A and r_B from B, with a positive charge q located at the midpoint.

Now, from Coulomb’s Law, the Force on q due to A

$$\begin{array}{l}{F}_{qA}=\frac{q}{4\pi {\epsilon }_0{r_A}^2}\text{towards}A\end{array}$$

Force of Q due to B

$$F_{qB}=\frac{1}{4\pi {\epsilon }_0}\frac{q}{r_B^2}\text{ towards }B$$

For the equilibrium of q, the two forces must be equal in magnitude and opposite in direction, i.e. they must be collinear.

|Fq_A| = |Fq_B|

$$\frac{1}{4\pi {\epsilon }_0}.\frac{q}{{r_A}^2}=\frac{1}{4\pi {\epsilon }_0}.\frac{q}{{r_B}^2}$$

Therefore, r_A = r_B.

For the equilibrium of q, it must be equidistant from A and B, i.e., in the middle of AB.

Now, for the equilibrium of the system, A and B must be in balance.

For the equilibrium of A,

Force on A by q $$\frac{1}{4\pi {\epsilon }_0}.\frac{q}{r_A^2}$$ towards q

Force on A by B $$=\frac{1}{4\pi {\epsilon }_0}.\frac{(1)(1)}{(r_A+r_B{)}^2}$$

$\begin{array}{l}=\frac{1}{4\pi {\epsilon }_0}.\frac{1}{{(2r_A)}^2}\end{array}$ away from q

For equilibrium to be achieved, the two forces must be equal, opposite and collinear. Therefore,

$\begin{array}{l}\frac{1}{4\pi {\epsilon }_0}.\frac{q}{{r_A}^2}=\frac{1}{4\pi {\epsilon }_0}.\frac{1}{{(2r_A)}^2}\end{array}$

or q = 1/4 in the magnitude of either charge.

For the equilibrium of B, it can be shown that the magnitude of q must be 1/4 of the magnitude of either charge.

Problem 3: Calculate the force between two positive charges, one with a charge of 6×10^-6C and the other with a charge of 4×10^-6C, when they are 0.040m apart.

Response:

• Q1 = 6 x 10^-6C
• Q2 = 4 x 10^-6C
• r = 0.040 m

$\begin{array}{l}{F}_e=k\frac{q_1{q}_2}{r^2}\end{array}$

$\begin{array}{l}{F}_e=1.07\times {10}^{-3};N\end{array}$

$\begin{array}{l}{F}_e=5.6\times {10}^{16}\end{array}$

$\begin{array}{l}{F}_e=13485\end{array}$

Fe = 134.85 N

Problem 4: What is the magnitude of the electric force between two point charges, q1 = +9 μC and q2 = 4 μC, separated by a distance of 12 cm?

Given:

k = 8.988 x 10⁹ Nm²C^-2

Q1 = 9 x 10^-6 C

Q2 = 4 × 10^-6 C

r = 0.12 m = 12 cm

Sol:

$F_e=k\frac{q_1{q}_2}{r^2}$

$\begin{array}{l}{F}_e=\frac{2.66\times {10}^{-9}}{0.0144}\end{array}$

$\begin{array}{l}{F}_e=6.25\times {10}^{20}\end{array}$

$\begin{array}{l}{F}_e=224.8611\end{array}$

F_e = 22.475 N