Coulomb's Law Class 12 Notes | EduRev

Physics Class 12

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COULOMB'S LAW
Charles Augustin de Coulomb, in 1785 through his experiments, found out that the two point charges 'q1' and 'q2' kept at a distance 'r' in a medium exert an electrostatic force 'F' on each other. The value of force F is given by
Coulomb`s Law Class 12 Notes | EduRev

This law gives the net electrostatic force experienced by q1 due to q2 and vice versa.
Coulomb`s Law Class 12 Notes | EduRev

Where,
F gives the magnitude of electrostatic force,
'q1 and 'q2 are the magnitudes of the two interacting point charges,
K is electrostatic constant which depends upon the medium surrounding the two charges. For vacuum, it is equal to 1/4πε0, the symbol ε0 is called ‘epsilon naught’ and it represents permittivity of vacuum.

K = 1/4πε0 = 9 × 109 Nm2/c2
ε0 = 8.854 × 10-12 C2/Nm2

If the point charges are kept in a medium with permittivity ε, then the electrostatic force between the point charges will be:
Coulomb`s Law Class 12 Notes | EduRev

This force F acts along the line joining the two charges and is repulsive if q1 and q2 are of the same sign and it is attractive if they are of opposite sign because like charges repel each other and unlike charges attract each other.

Refer following cases: 

Coulomb`s Law Class 12 Notes | EduRev

Limitations of Coulomb’s Law
Coulomb’s Law is derived under certain assumptions and can’t be used freely like other general formulas. The law is limited to following points:

  • We can use the formula if the charges are static ( in rest position)
  • The formula is easy to use while dealing with charges of regular and smooth shape, and it becomes too complex to deal with charges having irregular shapes
  • The formula is only valid when the solvent molecules between the particle are sufficiently larger than both the charges

SOLVED EXAMPLE
Q.1. Two charges 1 C and – 3 C are kept at a distance of 3 m. Find the force of attraction between them.
Ans. We have q1 = 1C, q2 = – 3C and r = 3m. Then using Coulomb’s Law and substituting above values we get 
F = k q1q2/ r2 
Or, F = 9 × 10× 1 × 3/ 32 
F = 3 × 109 Newton 

Q.2. The electron and proton of a hydrogen atom are separated (on the average) by a distance of approximately 5.3 × 10–11 m. Find the magnitudes of the electric force and the gravitational force between the two particles.
Ans. We can treat the electron and the proton of the hydrogen atoms as two point charges kept at a distance of 5.3 × 10–11 m from each other in vacuum. Also, we know that
Mass of an electron = me = 9.1 × 10-31 Kg
Mass of a proton = mp = 1.67 × 10-27 Kg
Charge of an electron = qe = - 1.6 × 10-19 C
Charge of a proton = qp = +1.6 × 10-19 C

Thus,
By Universal Law of Gravitation,
Coulomb`s Law Class 12 Notes | EduRev
 Coulomb`s Law Class 12 Notes | EduRev
By Coulomb’s Law,
Coulomb`s Law Class 12 Notes | EduRev
Coulomb`s Law Class 12 Notes | EduRev
= 8.2 × 10-8 N
Note: The massive difference between the magnitudes of gravitational force and electrostatic force. The electrostatic force between the electron and the proton of the hydrogen atom is around 1039 times stronger than the gravitational force between them. Thus, from now onwards we will consciously neglect the gravitational force of attraction between charged particles.

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