Understanding the Equation for Electric Force: Coulomb’s Law

The study of electrostatic interactions between charged particles is a cornerstone of classical physics. Coulomb’s Law is the mathematical equation that describes the electric force between charged objects. This article will delve into the equation for electric force, dissect its components, and explore its applications in various scientific and engineering contexts.

Coulomb’s Law: The Equation

The equation for electric force, known as Coulomb’s Law, is given by:

\[

F = k \frac{{|q_1 \cdot q_2|}}{{r^2}}

\]

Where:

– \( F \) is the magnitude of the electric force between the charges

– \( k \) is Coulomb’s constant, \(8.9875 \times 10^9 \, \text{N m}^2/\text{C}^2\)

– \( q_1 \) and \( q_2 \) are the magnitudes of the electric charges

– \( r \) is the distance between the centers of the two charges

Components Explained

1. Force (\( F \)): Measured in Newtons (N), this is the electric force experienced by the charges. The force is repulsive if the charges are of the same sign and attractive if the charges are of opposite signs.

2. Coulomb’s Constant (\( k \)): A fundamental constant that sets the scale for electric interactions. Its value has been precisely measured and is universally constant.

3. Charges (\( q_1 \) and \( q_2 \)): These are the magnitudes of the electric charges measured in Coulombs (C). Electric charge is a scalar quantity, and its sign determines the nature of the force (attraction or repulsion).

4. Distance (\( r \)): This is the distance between the centers of the two charges. It is measured in meters (m). The force decreases as the distance increases, following an inverse square law.

Applications

Electrical Engineering

In electrical circuits, understanding the electric force between charges can help in the design of electrical components like capacitors and resistors.

Medicine

Electrostatic forces are critical in various medical equipment like MRI machines and electrocardiograms.

Astronomy

The electric force plays a role in cosmic phenomena, although it is often overshadowed by gravitational forces due to the large distances involved.

Environmental Science

Understanding electrostatic interactions can help in pollution control. For instance, electrostatic precipitators use electric force to remove particles from exhaust gases.

Conclusion

Coulomb’s Law provides the mathematical framework to understand the forces between electric charges. Its applications range from the infinitesimally small scales in quantum mechanics to the vast expanses of cosmic interactions. Understanding this fundamental equation is key to advancing in fields as diverse as physics, engineering, and medicine.

**PROBLEMS AND SOLUTIONS**

Conceptual Questions

1. What is the effect on the electric force if the distance between two charges is halved?

Solution: According to Coulomb’s Law, the electric force is inversely proportional to the square of the distance (\(r^2\)). If the distance is halved, the force will increase by a factor of \( (1/2)^{-2} = 4 \).

2. What happens to the electric force if one of the charges is doubled?

Solution: If one of the charges is doubled, the electric force will also double. The force is directly proportional to the magnitudes of the charges \(q_1\) and \(q_2\).

3. Do like charges attract or repel each other?

Solution: Like charges repel each other.

4. Is Coulomb’s Law a vector equation?

Solution: Yes, Coulomb’s Law is inherently a vector equation, although the form often presented is for magnitude only.

5. What is the significance of Coulomb’s constant?

Solution: Coulomb’s constant sets the scale for electric interactions. It is a fundamental constant that describes how electric charges interact in vacuum.

6. Why does the electric force decrease as the distance between two charges increases?

Solution: According to Coulomb’s Law, electric force decreases as the square of the distance between the charges increases (\(1/r^2\)).

7. How does the electric force compare to the gravitational force between two electrons?

Solution: The electric force between two electrons is much stronger than the gravitational force between them.

8. Can electric force be negative?

Solution: The magnitude of electric force is always positive, but the vector direction can imply attraction (often considered as ‘negative’ force) or repulsion (‘positive’ force).

9. What happens to the electric force if the medium between the charges changes?

Solution: In a medium other than vacuum, the electric force is usually reduced. This is accounted for by a dielectric constant.

10. Is electric force a contact or non-contact force?

Solution: Electric force is a non-contact force, as it acts over a distance.

11. Calculate the electric force between two \(1 \, \text{C}\) charges separated by \(1 \, \text{m}\).

Solution: Using \( F = k \frac{{q_1 \cdot q_2}}{{r^2}} \), \( F = 8.9875 \times 10^9 \frac{{1 \times 1}}{{1^2}} = 8.9875 \times 10^9 \, \text{N} \).

2. Two charges of \(2 \, \text{C}\) and \(-3 \, \text{C}\) are \(0.5 \, \text{m}\) apart. What is the force between them?

Solution: \( F = 8.9875 \times 10^9 \frac{{|2 \times -3|}}{{0.5^2}} = -2.1594 \times 10^{10} \, \text{N} \) (The negative sign indicates attraction).

3. A charge of \(5 \, \text{C}\) experiences a force of \(10 \, \text{N}\) when placed \(2 \, \text{m}\) from another charge. Find the second charge.

Solution: \( q_2 = \frac{{F \times r^2}}{{k \times q_1}} = \frac{{10 \times 4}}{{8.9875 \times 10^9 \times 5}} = 8.89 \times 10^{-10} \, \text{C} \).

4. What is the force between two protons \(1 \times 10^{-15} \, \text{m}\) apart?

Solution: \( F = 8.9875 \times 10^9 \frac{{(1.6 \times 10^{-19})^2}}{{(1 \times 10^{-15})^2}} = 2.3 \, \text{N} \).

5. Two charges separated by \(10 \, \text{cm}\) experience a force of \(20 \, \text{N}\). If one charge is \(3 \, \text{C}\), find the other.

Solution: \( q_2 = \frac{{F \times r^2}}{{k \times q_1}} = \frac{{20 \times 0.1^2}}{{8.9875 \times 10^9 \times 3}} = 7.41 \times 10^{-10} \, \text{C} \).

6. What is the new force if the distance in Question 5 is doubled?

Solution: \( F_{\text{new}} = \frac{{F \times r^2}}{{4 \times r^2}} = \frac{{20}}{4} = 5 \, \text{N} \).

7. Two charges of \(+4 \, \text{C}\) and \(-4 \, \text{C}\) are separated by \(5 \, \text{m}\). What is the force?

Solution: \( F = 8.9875 \times 10^9 \frac{{|4 \times -4|}}{{5^2}} = -2.8776 \times 10^9 \, \text{N} \) (Negative sign indicates attraction).

8. What is the force between two electrons \(2 \, \text{m}\) apart?

Solution: \( F = 8.9875 \times 10^9 \frac{{(1.6 \times 10^{-19})^2}}{{2^2}} = 5.74 \times 10^{-10} \, \text{N} \).

9. Two charges separated by \(10 \, \text{m}\) exert a force of \(5 \, \text{N}\) on each other. What is the product \(q_1 \times q_2\)?

Solution: \( q_1 \times q_2 = \frac{{F \times r^2}}{k} = \frac{{5 \times 100}}{8.9875 \times 10^9} = 5.57 \times 10^{-9} \, \text{C}^2 \).

10. A \(2 \, \text{C}\) charge experiences a \(12 \, \text{N}\) force from another charge \(3 \, \text{m}\) away. What is the other charge?

Solution: \( q_2 = \frac

{{F \times r^2}}{{k \times q_1}} = \frac{{12 \times 9}}{8.9875 \times 10^9 \times 2} = 6 \times 10^{-10} \, \text{C} \).