Coulomb’s Law: Formula, Explanation, and Solved Problems

17.1.1 Electric Charge

1. Two kinds of charges

There are two types of charges:

• Positive charge
• Negative charge

2. Quantity of charge

The quantity of charge carried by an object is measured in coulomb (C).

Example:
An object carries an excess charge of +2 μC or –3 μC.

17.1.2 Point Charge

1. A charge carried by a particle whose size is very small is known as a point charge.

Examples:

• An electron is a very tiny particle (according to the old model) and it carries a negative charge e where
  
                                             $e=-1.6 \times 10^{-19} \ C$
  
  
• A proton in an atom is also a very tiny particle. It carries a charge of +e.
  
  

2. A large charged object may be considered to be a point charge when we look at it from a distance that is very large compared with its size.

17.1.3 Neutral and Charged Objects

1. Neutral object

Any object carries equal amount of potitve charge and negative charge is electrically neutral.

2. Charged object

(a) A neutral object becomes negatively charged when it acquires some extra electrons or negative ions.
(b) A neutral object becomes positively charged when it loses some electrons or acquires some positively charged ions.
(c) Hence, a neutral object becomes charged when it carries an excess negative or positive charge.

3. Excess charge in terms of e

The excess charge Q carried by a charged object may be expressed as a multiple of the electronic charge e, as follows:

$Q= ne$

where n is the number of electrons acquired (resulting in the object carrying an excess negative charge) or lost (resulting in the object carrying an excess positive charge) by the object.

17.1.4 Electric Force

The electric force has the following features:
(a) An electric force exists between two charges.
(b) (i). Unlike charges attract
      (ii) Like charges repel

17.1.5 Coulomb’s Law

1. Consider two stasionary point charges $Q_1$ and $Q_2$ in vacuum separated by a distance r. Then an electric force of magnitude F act on each charge, as shown in Figure 17.1

2. The two forces form an action-reaction pair. Hence, according to Newton's third law of motions, the magnitude of the force  acting on $Q_1$ must be equal to the magnitude of the force acting on $Q_2$.

3. The magnitude of the force in free space is determined by the following quantities:

(a) $F \propto Q_1$

(b) $F \propto Q_2$

(c) $F \propto \frac{1}{r^2}$

Hence, we have

$F \propto \frac{Q_1Q_2}{r^2}$  

$F =k \frac{Q_1Q_2}{r^2}$ (in free space)

where k is a constant if $Q_1$ and $Q_2$ are in free space.

The constant k is given by

$k=\frac{1}{4 \pi \epsilon_0}$

Hence, we have

$F =k \frac{Q_1Q_2}{r^2}$ (in free space)

This equatins is known as Coulomb's law. (Note: The equation is only used to determine the electric force acting on two static point charges.)

17.1.6 The Constant $\epsilon_0$

1. Definition

$\epsilon_0$ is known as the permittivity of free space

2. Value and unit

$\epsilon_0$ has a value of

$\epsilon = 8.85 \times 10^{-12} \ F.m^{-1}$

where F represents farad (unit for capacitance).

3. Value of k

$k=\frac{1}{4\pi \epsilon_0}$

$k=\frac{1}{4 \pi \times 8.85 \times 10^{-12}}=8.99 \times 10^9$

$k \approx 9.0 \times 10^9 \ F^{-1}.m$

4. In Coulomb's law

We have

Therefore, $k = 9.0 \times 10^{9} \ N.m^2.C^{-2}$

$F =k \frac{Q_1Q_2}{r^2}$

$F \approx (9.0 \times 10^9) \frac{Q_1Q_2}{r^2}$ (in free space)

Example 17.1

An atom was neutral before it becomes negatively charged when it has acquired electrons. If it carries an excess charge of –3.2 μC, determine the number of electrons it has acquired.

Solution

$n = \frac{Q}{e}$

$n = \frac{-3.2 \times 10^{-6} \ C}{-1.6 \times 10^{-19} \ C}$

$n=2 \times 10^{13}$

Example 17.2

Three point charges are firmly held on a straight line of length 4.0 cm, as shown in Figure 17.2. Dtermine the resultant electric force acting on (a) charge $q_2=+5.0 \mu C$ and (b) charge $q_1=+10 \mu C$.

Answer

(a) Force acting on $q_2$ due to $q_1$ is a repulsive force given by

$F_{12} \cong (9.0 \times 10^9) \frac{q_1q_2}{r^2}$

$F_{12} = (9.0 \times 10^9) \frac{(10 \times 10^{-6})(5 \times 10^{-6})}{(0.02)^2}$

$F_{12} = 1.13 \ kN$ ti the right.

Force acting on $q_2$ due to $q_3$ is a repulsive force given by

$F_{23} \cong (9.0 \times 10^9) \frac{q_2q_3}{r^2}$

$F_{23} = (9.0 \times 10^9) \frac{(5 \times 10^{-6})(8 \times 10^{-6})}{(0.02)^2}$

$F_{23} = 0.90 \ kN$ ti the right.

Resultant force acting on $q_2$ is

$F = F_{12} + F_{23}$

$F=(+1.13) + (+0.90) = +2.03 \ kN$ to the right

(b) Force acting on $q_1$ due to $q_2$ is a repulsive force given by

$F_{12} = 1.13 \ kN$ ti the left.

Force acting on $q_1$ due to $q_3$ is a repulsive force given by

$F_{13} \cong (9.0 \times 10^9) \frac{q_1q_3}{r^2}$

$F_{13} = (9.0 \times 10^9) \frac{(10 \times 10^{-6})(8 \times 10^{-6})}{(0.02)^2}$

$F_{13} = 0.45 \ kN$ ti the right.

Resultant force acting on $q_2$ is

$F = F_{12} + F_{13}$

$F=(-1.13) + (+0.90) = 0.68 \ kN$ to the left

Example 17.3

Two small light balls, each of mass 2.0 g and carrying equal amount of like charge, are suspended freely by light non-conducting strings. The length of each string is 13 cm. The balls are in static equilibrium and separated by a distance of 10 cm, as shown in Figure 17.3.

(a) Draw a free body force diagram to show all the external forces that act on one ball.

(b) Determine the charge carried by each ball.

Answer

(a) W = weight of ball = mg

$F_1$ = tension in string

$F_2$ = repulsive electric force

(b) Refer to the vector diagram shown in Figure 17.4(b)

$F_2=W \ tan \theta$

But $\tan \theta = \frac{5}{\sqrt{13^2-5^2}}=\frac{5}{12}$

$F_2=mg \ tan \theta$

$F_2=(0.002)(9.8) \frac{5}{12}=8.17 \times 10^{-3}$ N

17.1.7 Applications

Coulomb’s Law describes the electric force between two point charges. 

The force depends on the magnitude of the charges and the distance between them.

Main Applications 

1. Interaction Between Electric Charges

Coulomb’s law is used to calculate the force between two charged particles such as electrons or protons.

Example:

• Electron–electron interaction

• Proton–electron interaction in atoms

This helps explain atomic structure and electrostatic forces in matter.

2. Atomic and Molecular Physics

In atoms, Coulomb’s law explains the attraction between the nucleus and electrons.

Applications:

• Structure of atoms

• Bonding between atoms

• Molecular forces

For example, the attraction between a proton and an electron keeps the electron orbiting the nucleus.

3. Electrostatic Devices

Many devices work based on electrostatic forces predicted by Coulomb’s law.

Examples:

• Photocopiers

• Laser printers

• Electrostatic precipitators (remove dust from factory smoke)

These devices use charged particles that attract or repel each other.

4. Electric Field Calculations

Coulomb’s law helps determine the electric field produced by charges.

Applications include:

• Electric field around charged conductors

• Field around charged particles

• Capacitor design

5. Charging of Objects

It explains phenomena such as:

• Static electricity

• Charging by friction

• Attraction of small objects (paper pieces) by a charged comb or ruler.

Example:
When a plastic comb is rubbed with hair, it becomes charged and can attract small paper pieces.

6. Particle Physics and Plasma Physics

Coulomb’s law is used to analyze:

• Motion of charged particles

• Behavior of plasma

• Particle accelerators

17.1.8 Frequently Asked Questions (FAQ) 

1. What is Coulomb’s Law?

Coulomb’s Law describes the electric force between two charged objects. The force depends on the magnitude of the charges and the distance between them.

$F =k \frac{Q_1Q_2}{r^2}$

Where: 

• F = electric force (N)

• Q₁, Q₂ = charges (C)

• r = distance between charges (m)

• k = Coulomb’s constant $\mathrm{\$9\; \backslash times\; 10^9\; \backslash \; N.m^2.\; C^\{-2\}\$}$

2. Who discovered Coulomb’s Law?

Coulomb’s Law was discovered by the French physicist Charles-Augustin de Coulomb in 1785 using a device called a torsion balance.

3. What happens if the distance between charges increases?

The electric force decreases rapidly because the force is inversely proportional to the square of the distance.

Example:

• If the distance becomes 2 times larger, the force becomes 1/4 of the original force.

4. What happens if the charges become larger?

The electric force becomes stronger because the force is directly proportional to the magnitude of the charges.

Example:

• If one charge doubles, the force doubles.

• If both charges double, the force becomes four times larger.

5. Do like charges attract or repel?

• Like charges repel ( + with + , − with − )

• Unlike charges attract ( + with − )

6. Where is Coulomb’s Law used?

Coulomb’s Law is applied in many areas of physics, such as:

• Atomic physics

• Electrostatic devices (photocopiers, laser printers)

• Electric field calculations

• Particle physics

17.1.9 Conclusion  

Coulomb’s Law is a fundamental principle in electrostatics that explains how electric charges interact with each other. It states that the electric force between two charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.

From this law, we understand several important concepts:

• Like charges repel, while unlike charges attract.

• The greater the charge, the stronger the electric force.

• The greater the distance between charges, the weaker the force.

• The interaction occurs along the line connecting the two charges.

Coulomb’s Law is essential for understanding many phenomena in physics, including electric fields, electric potential, atomic structure, and the behavior of charged particles. It also forms the basis for many technologies involving electrostatics, such as capacitors, photocopiers, laser printers, and electronic devices.

In summary, Coulomb’s Law provides a clear mathematical description of how electric charges influence each other, making it one of the cornerstones of classical electromagnetism.

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