Coulomb's Law is a cornerstone of electrostatics, a branch of physics that deals with electric charges at rest. It provides a mathematical expression for the force between two electrically charged particles and plays a crucial role in understanding the behavior of electric charges. In Class XII, students embark on an exploration of this fundamental law, uncovering the mysteries of electrostatic forces and their applications. This article aims to shed light on Coulomb's Law and its significance in the world of electromagnetism.
The Discovery of Coulomb's Law:
The law is named after Charles-Augustin de Coulomb, a French physicist, who first formulated it in the 18th century. Coulomb conducted meticulous experiments to investigate the force between two electrically charged objects. His groundbreaking work laid the foundation for our understanding of electrostatics and paved the way for future advancements in the field.
The Statement of Coulomb's Law:
Coulomb's Law states that the electrostatic force (F) between two point charges (q₁ and q₂) is directly proportional to the product of their magnitudes and inversely proportional to the square of the distance (r) between them. Mathematically, it can be expressed as:
F = k * (|q₁ * q₂|) / r²
Here, k is Coulomb's constant, a fundamental constant of nature that depends on the medium between the charges and the units used. It acts as a scaling factor to ensure the force is measured in Newtons (N).
Applications of Coulomb's Law:
Coulomb's Law finds numerous practical applications in our daily lives and advanced technologies. For instance:
It explains the interaction between charged particles in electrical circuits, influencing current flow and device operation.
In electronic devices, the law governs the attraction and repulsion of charges, enabling the creation of sophisticated microchips and transistors.
Understanding Coulomb's Law is essential in designing electrical equipment, such as generators, motors, and transformers, that rely on electromagnetic interactions.
It is also the basis for developing systems to measure charge and voltage, like the electroscope and voltmeter.
Conclusion:
Coulomb's Law stands as a fundamental pillar in the realm of electrostatics. As Class XII students delve into the world of electromagnetism, they encounter this essential law that governs the forces between electric charges. By understanding Coulomb's Law, students gain insights into the intricacies of electrical interactions, paving the way for technological innovations that shape our modern society. As we continue to harness the power of electric forces, a firm grasp of Coulomb's Law will undoubtedly illuminate the path to new discoveries and applications in the ever-expanding field of electromagnetism.
The Nature of Electric Force:
The law reveals that like charges (both positive or both negative) repel each other, while opposite charges (one positive and one negative) attract each other. This attraction and repulsion create the foundation for various phenomena, such as the behavior of magnets, the functioning of electronic devices, and the structure of atoms and molecules.
Significance of Distance:
Coulomb's Law also emphasizes the crucial role of distance in determining the strength of the electric force. As the distance between two charges increases, the force between them decreases rapidly. Inversely, bringing the charges closer intensifies the force. This relationship underscores the importance of maintaining safe distances in electrical systems and power transmission.
FAQs on Coulomb's Law
Q: Two insulated charged copper spheres A and B have their centers separated by a distance of 50 cm. What is the mutual force of electrostatic repulsion if the charge on each is 6.5 × ?0^−? C? The radii of A and B are negligible compared to the distance of separation. (b) What is the force of repulsion if each sphere is charged double the above amount, and the distance between them is halved?
A: Charge on sphere A, qA = 6.5 × 10-7 C
Charge on sphere B, qB = 6.5 × 10-7 C
Distance between the spheres, r = 50 cm = 0.5 m
Force of repulsion between two spheres
F = 1/4??0× q1q2 / r2
, where ?0 = Permittivity of free space = 8.854 × 10-12 C2N-1m-2
Therefore, F = 1 / 4×?×8.854 ×10-12 × 6.5 ×10-7×6.5 ×10-7 / 0.52 N = 0.0152 N = 1.52 × 10-2 N
(b) Charge on sphere A, qA = 2 × 6.5 × 10-7 C = 1.3 × 10-6 ?
Charge on sphere B, qB = 2 × 6.5 × 10-7 C = 1.3 × 10-6 ?
Distance between the spheres, r = 50 / 2 = 25 cm = 0.25 m
Force of repulsion between two spheres
F = 1/4??0× q1q2 / r2 , where ?0 = Permittivity of free space = 8.854 × 10-12 C2N-1m-2
Therefore, F = 1 / 4×?×8.854 ×10-12 × 1.3 ×10-6 × 1.3 ×10-6 / 0.252 N = 0.243 N
Q: Suppose the spheres A and B in Exercise 1.12 have identical sizes. A third sphere of the same size but uncharged is brought in contact with the first, then brought in contact with the second, and finally removed from both. What is the new force of repulsion between A and B?
A: When the third uncharged sphere C is brought in contact with the sphere A, then the charge is shared and becomes half. Then
qA = ?/2 and qC = ?/2
When the charged sphere C is brought in contact with charged sphere B, the charge between
both the sphere is shared and becomes half
qB,qC = 1/2 (q + ?/2) = 3?/4
Hence the force of repulsion between sphere A and B can be given as
F = 1/4??0× q1q2 / r2 , where ?0 = Permittivity of free space = 8.854 × 10-12 C2N-1m-2
= 1 / 4×?×8.854 ×10-12 X (q/2 X 3q/4) /0.52 = 1 / 4×?×8.854 ×10-12 X 3 X q2 / 8X0.52
= 1 / 4×?×8.854 ×10-12 X 3 X (6.5 × 10-7)2 / 8X0.52 = 5.695 × 10-3 N
Q: Figure 1.33 shows tracks of three charged particles in a uniform electrostatic field. Give the signs of the three charges. Which particle has the highest charge to mass ratio?
A: Since the charges 1 & 2 are attracted towards + ve, their charges will be – ve. The charge 3 is attracted towards – ve, hence its charge will be +ve.
The charge to mass ratio (emf) is directly proportional to the displacement, charge 3 will have the highest charge to mass ratio.
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