Coulomb's Law

Coulomb's Law

• The first quantitative experiments on electric charge forces were by Charles-Augustin Coulomb (1736-1806).
• Coulomb measured the attractive and repulsive forces between charges and formulated Coulomb's Law.
• The electrical force exerted by one charged body on another is:
  • Directly proportional to the product of the magnitudes of the two charges
  • Inversely proportional to the square of the distance between their centers
• F ∝ |q1q2| / r²

Magnitude of Mutual Force

• F is the magnitude of the mutual force acting on the two charges q1 and q2.
• r is the distance between the centers of the charges.
• The force on each charge acts along the line connecting the charges.
• The force exerted by q1 on q2 is equal in magnitude but opposite in direction to the force exerted by q2 on q1 (Newton's third law).
• F = K(q1q2/r²) where K is the Coulomb constant.
• Coulomb's law generally holds for charged objects much smaller than the distance between them (point charges).
• Coulomb's law resembles Newton's inverse square law of gravitation (F = Gm1m2/r²).
• Both laws follow the inverse square law.
• Charge (q) in Coulomb's law is analogous to mass (m) in Newton's law.
• Gravitational forces are always attractive, while electrostatic forces can be repulsive or attractive, depending on the signs of the charges.
• K = 1/(4πε0), ε0 is the electric constant (permittivity).
• ε0 ≈ 8.854 x 10^-12 C²/N·m².
• K ≈ 8.99 x 10^9 N·m²/C².
• Coulomb's law can be written as F = (1/(4πε0)) * (|q1q2| / r²).
• With K in SI units, 'q' is in coulombs and 'r' is in meters, 'F' results in newtons.

Vector Form of Coulomb's Law

• Charges q1 and q2 are separated by a distance r.
• If charges have the same sign, the force is repulsive.
• If charges have opposite signs, the force is attractive.
• The force on particle 1 exerted by particle 2 is F12.
• When charges have the same sign, F12 is parallel to r12.
• When charges have opposite signs, F12 is anti-parallel to r12.
• F = F12 + F13 + ... (Superposition principle for multiple charges).
• F12 is the force on particle 1 from particle 2.
• F13 is the force on particle 1 from particle 3.

Significance of Coulomb's Law

• Coulomb's law describes forces exerted by charged spheres.
• When incorporated into quantum physics, this law accurately describes:
  • Electrical forces binding electrons to an atom's nucleus
  • Forces binding atoms to form molecules
  • Forces binding atoms and molecules to form solids or liquids
• Most daily experience forces (excluding gravitational) are electrical.

Electric Field Intensity (E)

• Electric field intensity (E) is the electric force experienced by a unit positive test charge placed at a point.
• To find E at a point P, place a small positive test charge q0 at P.
• Determine the force F experienced by q0 due to other charges.
• E = F/q0
• The direction of E is the same as the direction of F.
• The SI unit of electric field intensity is N/C (newtons per coulomb).

Electric Field Intensity Due to a Point Charge

• To find E at a distance r from a point charge q, place a test charge q0 at that point.
• Force applied on q0 by q: F = (1/(4πε0)) * (qq0/r²) * r̂
• Electric field intensity is: E = (1/(4πε0)) * (q/r²) * r̂
• The direction of E is along the radial line from q:
  • Outward if q is positive
  • Inward if q is negative
• Magnitude of E: E = (1/(4πε0)) * (q/r²)

Electric Field of System of N-Charges

• With N charges q1, q2, q3, ..., qN at positions r1, r2, r3, ..., rN relative to a point P.
• The electric field E at P is the vector sum of the electric intensities due to each charge:
• E = E1 + E2 + ... + EN
• E = (1/(4πε0)) * Σ(qi/ri²) * r̂i for i = 1 to N.

Electric Field Due to Continuous Charge Distribution

• Electric charge is quantized; large collections are continuous charge distributions.
• Types of continuous charge distributions:
  • Linear Charge Distribution
  • Surface Charge Distribution
  • Volume Charge Distribution

Electric Field Due to Linear Charge Distribution

• Positive source charge distributed uniformly along a line.
• Small charge element dq with length ds.
• Linear charge density: λ = dq/ds => dq = λds
• Total charge: q = ∫λds
• Electric field: E = (1/(4πε0)) * ∫(λds/r²) * r̂

Electric Field Due to Surface Charge Distribution

• Positive source charge distributed uniformly over a surface.
• Small area element dA with charge dq.
• Surface charge density: σ = dq/dA => dq = σdA
• Total charge: q = ∫σdA
• Electric field: E = (1/(4πε0)) * ∫(σdA/r²) * r̂

Electric Field Due to Volume Charge Distribution

• Positive source charges distributed uniformly in a volume V.
• Small volume element dV with charge dq.
• Volume charge density: ρ = dq/dV => dq = ρdV
• Total charge: q = ∫ρdV
• Electric field: E = (1/(4πε0)) * ∫(ρdV/r²) * r̂

The Electric Dipole

• Two equal and opposite charges separated by a small distance form a dipole.
• Consider charges +q and -q separated by a distance d along the z-axis.
• Electric field is calculated at point P at a distance x along the perpendicular bisector.
• The positive charge (+q) and negative charge (-q) setup an electric field at point P.
• The distance to point P from +q and -q is the same.
• The magnitudes of the electric fields created by each charge are:
  • E1 = (1/(4πε0)) * (q/r²)
  • E2 = (1/(4πε0)) * (q/r²)
• Resolving the electric fields into components:
  • E1z = E1cosθ
  • E1x = E1sinθ
  • E2z = E2cosθ
  • E2x = E2sinθ
• E1z and E2z are equal in magnitude and cancel out.
• E1x and E2x are equal in magnitude and add to give the resultant electric field.
• Total magnitude of electric intensity at P: E = E1x + E2x = E1cosθ + E2cosθ
• E = 2E1cosθ, since E1 = E2 = E
• E = (1/(4πε0)) * (q/r²) and cosθ = (d/2)/r
• Using r = √(x² + (d/2)²),
• E = (1/(4πε0)) * (qd/r³)
• E = (qd/(4πε0x³)) * [1 - (3/2)(d/2x)²] obtained after binomial expansion
• Electric dipole moment: P = qd
• It is directed from the negative to the positive charge.
• E = (1/(4πε0)) * (P/x³)