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COULOMB'S LAW

  • Topic Overview

    • The study of the electric force between point charges, defined as Coulomb's Law.

Objectives

  • State Coulomb’s Law.

  • Explain the relation between electric force and the magnitude and distance of point charges.

  • Solve problems involving electrostatic forces using Coulomb’s Law.

Behavior of Charges

  • Like Charges:

    • Like charges repel each other (e.g., + with +).

  • Opposite Charges:

    • Opposite charges attract each other (e.g., + with -).

Charles Augustine De Coulomb

  • Relationship to Coulomb's Law:

    • Devoted his life to studying charges.

    • Discovered a method to calculate the electric force of a point charge using Coulomb’s torsion balance.

    • Unit of charge: Coulombs (C), named after him.

    • Introduced the Coulomb’s Law and its equation.

Coulomb’s Law / Equation

  • Formula: F=kq<em>1q</em>2r2F = k \frac{|q<em>1 q</em>2|}{r^2}

    • Where:

    • F: Electric Force (N)

    • q1: Point charge 1 (C)

    • q2: Point charge 2 (C)

    • r: Distance between the two charges (m)

    • k: Coulomb’s constant (approximately 9.0×109 Nm2/C29.0 \times 10^9 \text{ Nm}^2/\text{C}^2).

Relation of Electric Force to Magnitude and Distance

  • Proportionality:

    • Electric Force is directly proportional to the magnitude of the point charges and inversely proportional to the distance between the point charges.

Example Exercises

  • Exercise 1:

    • Two objects, charges +1.0 C and -1.0 C, are 1.0 km apart.

    • Calculation:

    • Magnitude of attractive force: F=9.0×103NF = 9.0 \times 10^3 N.

  • Exercise 2:

    • A charge of +10μC is 25 cm from -20μC.

    • Calculation:

    • Electric force magnitude: F=30NF = 30 N.

  • Exercise 3:

    • Electric force between +800nC and +900nC is 15.0 N.

    • Calculation:

    • Distance: r=0.02mextor2×102mr = 0.02m ext{ or } 2\times 10^{-2} m.

  • Exercise 4:

    • Electrostatic force between 2.0 C and -1.0 C separated by 1.0 m:

    • Calculation:

    • Force: F=1.8×1010NF = 1.8 \times 10^{10} N.

  • Exercise 5:

    • Charges of 2 × 10^-7 C and 4.5 × 10^-7 C acted on by 0.1 N.

    • Calculation:

    • Distance: r=9×102mr = 9 \times 10^{-2} m.

  • Exercise 6:

    • Identical charges with electrostatic force of 1000 N, separated by 0.01 m.

    • Calculation:

    • Charge magnitude: 3×106Cextor3µC3 \times 10^{-6}C ext{ or } 3µC.

ELECTRIC FIELD

Introduction to Electric Field

  • Definition:

    • An electric field is a region around charged particles where other charged particles feel a force.

    • Denoted by E, it causes any charge within it to experience an electric force.

    • Every charge has an electric field associated with it.

Electric Field Lines of Force

  • Definition of Electric Field Lines:

    • Imaginary lines drawn to show the direction of the electric field at any point; tangent at any point represents the electric field vector.

Properties of Electric Field Lines

  • Direction:

    • Away from positive charges and toward negative charges.

  • Starting and Ending Points:

    • Start from positively charged particles and end on negatively charged particles or extend towards infinity.

  • Line Behavior:

    • Field lines do not intersect or break.

  • Density:

    • Field strength is indicated by the number of lines; closer lines indicate stronger fields.

Measuring Electric Field

  • Test Charge Method:

    • Electric field (E) is defined as the force experienced by a test charge situated at that point.

    • Formula:
      E=Fq0E = \frac{F}{q_0}

    • Where:

      • E: Electric Field (N/C)

      • F: Electric Force

      • q0: Test charge.

  • Coulomb's Calculation for Electric Field:

    • E=kqr2E = k \frac{q}{r^2}

    • Where q is the point charge, r is the distance from the charge.

EXERCISES

  • Exercise on Electric Field:

    • Calculate electric fields resulting from point charges using previously discussed values and equations.

Electric Flux

  • Definition:

    • Electric flux indicates the amount of electric field lines passing through a surface.

    • It is calculated as the dot product of the electric field and the area:
      ΦE=EimesAimescos(Θ)\Phi_E = E imes A imes \cos(\Theta)

    • Where:

      • Φ_E: Electric Flux (Nm²/C)

      • E: Electric Field (Nm)

      • A: Surface Area (m²)

      • Θ: Angle between the electric field and area vector.

Applications of Electric Flux in Practical Scenarios

  • Determine the electric flux through surfaces in various orientations to electric fields.

GAUSS'S LAW

  • Introduction:

    • Introduced by Carl Friedrich Gauss in 1835, relates electric fields at points on a closed surface to the net charge enclosed.

Gauss's Law Equation

  • Formula for Gauss's Law: Φ<em>E=qϵ</em>0\Phi<em>E = \frac{q}{\epsilon</em>0}

    • Where ε₀ is the permittivity of free space (≈ 8.85 x 10^-12 C²/Nm²).

Exercises on Gauss's Law

  1. Calculate electric flux for various charge configurations using the established equations.

Conclusion

  • The concepts of Coulomb’s Law, Electric Fields, and Gauss's Law are foundational to understanding electrostatics, providing insight into both theoretical calculations and practical applications in electrical engineering and physics.