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Coulomb’s Law

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Estimated time: 29 minutes
CBSE: Class 12

Introduction to Coulomb's Law

  • The electric interaction between two charged bodies can be expressed in terms of the forces they exert on each other.
  • Coulomb (1736–1806) made the first quantitative investigation of the force between electric charges, using point charges at rest.
  • A point charge is a charge whose dimensions are negligibly small compared to its distance from other bodies.
  • Coulomb's Law is a fundamental law governing the interaction between stationary charges.
CISCE: Class 12

Definition: Electric Force

The force of attraction or repulsion acting between two electric charges is called the electric force.

Maharashtra State Board: Class 11

Definition: Relative Permittivity (Dielectric Constant)

The ratio of the force between two point charges placed at a certain distance apart in free space or vacuum to the force between the same two point charges when placed at the same distance in the given medium is called relative permittivity (K) or dielectric constant.

CBSE: Class 12

Statement of Coulomb's Law

  • When the linear size of charged bodies is much smaller than the distance separating them, they are treated as point charges.
  • Coulomb found that the force between two point charges varied inversely as the square of the distance between them.
  • The force was directly proportional to the product of the magnitudes of the two charges.
  • The force acted along the line joining the two charges.
  • For two point charges q1​ and q2​ separated by a distance r in vacuum, the magnitude of the force is given by:
    F = k\[\frac {∣q_1q_2∣}{r^2}\] ...(1)
CBSE: Class 12
Maharashtra State Board: Class 11

Law: Coulomb's Law

The force of attraction or repulsion between two point charges at rest is directly proportional to the product of the magnitude of the charges and inversely proportional to the square of the distance between them.

Scalar Form:

F = \[\frac{1}{4\pi K\varepsilon_0}\cdot\frac{q_1q_2}{r^2}\]

Vector Form:

\[\vec F_{12}\] = \[\frac{1}{4\pi K\varepsilon_{0}}\cdot\frac{q_{1}q_{2}}{r_{12}^{2}}\hat{r}_{12}\]

where q1​ and q2 are charges separated by distance r, and \[\hat r_{12}\] is the unit vector from q1​ to q2​.

CBSE: Class 12

About Charles Augustin de Coulomb

  • Coulomb was a French physicist who began his career as a military engineer in the West Indies.
  • In 1776, he returned to Paris and retired to a small estate to pursue scientific research.
  • He invented a torsion balance to measure the quantity of a force and used it to determine forces of electric attraction or repulsion between small charged spheres.
  • In 1785, he arrived at the inverse square law, now known as Coulomb's Law.
  • The law had been anticipated by Priestley and also by Cavendish, though Cavendish never published his results.
  • Coulomb also found the inverse square law of force between unlike and like magnetic poles.
CBSE: Class 12

Experimental Verification of Coulomb’s Law

  • Coulomb used a torsion balance to measure the force between two charged metallic spheres.
  • When the separation between two spheres is much larger than their radius, the spheres may be regarded as point charges.
  • To vary the charge in a controlled way, Coulomb touched a charged metallic sphere (charge q) with an identical uncharged sphere — by symmetry, the charge on each became q/2.
  • Repeating this process gave charges of q/2, q/4, and so on.
  • He varied the distance for a fixed pair of charges and measured the force for different separations.
  • He then varied the charges in pairs, keeping the distance fixed, and compared forces for different pairs at different distances.
  • Through this method, Coulomb arrived at the relation in Eq. (1).
  • While originally established at a macroscopic scale, the law has also been confirmed down to the subatomic level (r ∼ 10−10 m).
CBSE: Class 12

The Constant k and the Unit of Charge

  • In Eq. (1), k is an arbitrary positive constant whose choice determines the size of the unit of charge.
  • In SI units, the value of k is approximately 9 × 109 Nm2C−2.
  • For q1 = q2 = 1 C and r = 1 m, the force works out to F = 9 × 109 N.
  • Therefore, 1 coulomb is the charge that, when placed 1 m from another charge of the same magnitude in vacuum, experiences a repulsive force of 9 × 109 N.
  • One coulomb is too large a unit for practical use; in electrostatics, smaller units like 1 mC or 1 μC are used instead.
  • The constant k is written as k = \[\frac {1}{4πε_0}\] for later convenience, giving:
    F = \[\frac {1}{4πε_0}\] ⋅ \[\frac {∣q_1q_2∣}{r^2}\] ...(2)
  • ε0​ is called the permittivity of free space, with a value of ε0 = 8.854 × 10−12 C2N−1m−2
CBSE: Class 12

Law: Vector Form of Coulomb's Law

  • Since force is a vector, Coulomb's Law is better expressed in vector notation.
  • The vector leading from charge 1 to charge 2 is r21 = r2 − r1, and from charge 2 to charge 1 is r12 = r1 − r2 = −r21.
  • The corresponding unit vectors are \[\hat r_{21}\] = \[\frac {r_{21}}{r_{21}}\]​​ and \[\hat r_{12}\] = \[\frac {r_{12}}{r_{12}}\]​​, with \[\hat r_{21}\] =−\[\hat r_{12}\]​.
  • The vector form of Coulomb's Law is:
    F21 = \[\frac {1}{4πε_0}\] ⋅ \[\frac {q_1q_2}{r^2_{21}}\]\[\hat r_{21}\] ...(3)
  • If q1​ and q2​ are of the same sign, F21​ is along \[\hat r_{21}\]​, representing repulsion.
  • If q1 and q2​ are of opposite signs, F21​ is along −\[\hat r_{21}\]​, representing attraction.
  • Eq. (3) handles both like and unlike charges correctly within a single equation — no separate formulas are needed.
  • The force F12​ on q1​ due to q2​ is obtained by interchanging 1 and 2: F12 = −F21, confirming agreement with Newton's Third Law.
  • Eq. (3) gives the force in vacuum; when charges are placed in matter, the situation becomes more complex due to the charged constituents of the medium.
CBSE: Class 12

Example 1

  • Both Coulomb's Law and Newton's Law of Gravitation have an inverse-square dependence on the distance between charges/masses, respectively.
  • The electric force between an electron and a proton at distance r is Fe = −\[\frac{1}{4\pi\varepsilon_0}\cdot\frac{e^2}{r^2}\], where the negative sign indicates attraction.
  • The gravitational force between them is FG = −G\[\frac{m_pm_e}{r^2}\], which is also always attractive.
  • The ratio of their magnitudes for an electron–proton pair is ∣\[\frac {F_e}{F_G}\]∣ = \[\frac {e^2}{4πε_0Gm_pm_e}\] = 2.4 × 1039.
  • For two protons, the ratio is ∣\[\frac {F_e}{F_G}\]∣ = \[\frac {e^2}{4πε_0Gm_p^2}\] = 1.3 × 1036​.
  • For two protons inside a nucleus (r ∼ 10−15 m), the electric force is Fe ∼ 230 N while the gravitational force is only FG ∼ 1.9 × 10−34 N.
  • These values show that electrical forces are enormously stronger than gravitational forces.
  • The magnitude of the electric force between an electron and a proton at r = 1 Å is ∣F∣ = 2.3 × 10−8 N.
  • The resulting acceleration of the electron is ae = 2.5 × 1022 m/s2, and that of the proton is ap = 1.4 × 1019 m/s2.
  • These enormous values confirm that the effect of the gravitational field is negligible on the motion of the electron under the Coulomb force.
CBSE: Class 12

Example 2

  • Sphere A carries charge q and sphere B carries charge q′, separated by r = 10 cm; the initial electrostatic force between them is F = \[\frac {1}{4πε_0}\] ⋅ \[\frac {qq′}{r^2}\]​.
  • When identical uncharged sphere C touches A, the charge redistributes equally, leaving each with q/2 by symmetry.
  • When identical uncharged sphere D touches B, the redistributed charge on each is q′/2.
  • When B is brought closer to A such that the new separation is r/2 = 5 cm, the new force is:
    F′ = \[\frac {1}{4πε_0}\] ⋅ \[\frac{(q/2)(q^\prime/2)}{(r/2)^2}=\frac{1}{4\pi\varepsilon_0}\cdot\frac{qq^\prime}{r^2}\] = F
  • The electrostatic force on A due to B remains exactly unchanged — halving the charges and halving the distance perfectly cancel each other out.
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