Lecture on Electric Charge Problems

Introduction

• Focus on solving problems associated with electric charge.

Problem 1: Charge of a Metal Rod

• Given: 4500 electrons transferred to a neutral metal rod.
• Charge of Electron:
  • One electron has a charge of -1.602 × 10⁻¹⁹ C.
• Solution:
  • Convert electrons to coulombs.
  • Calculation: 4500 electrons × (-1.602 × 10⁻¹⁹ C/electron) = -7.2 × 10⁻¹⁶ C.
  • Result: The metal rod acquires a charge of -7.2 × 10⁻¹⁶ C.

Problem 2: Charge of Moles of Electrons

• Given: 3.6 moles of electrons.
• Avogadro's Number: 1 mole = 6.022 × 10²³ electrons.
• Solution:
  • Convert moles to number of electrons.
  • Convert electrons to coulombs.
  • Calculation: 3.6 moles × 6.022 × 10²³ electrons/mole × -1.602 × 10⁻¹⁹ C/electron = -347,000.784 C.
  • Convert to kilocoulombs: -347 kC.
  • Result: Electric charge is approximately -347 kC.

Problem 3: Charge of Protons

• Given: 50 kg of protons.
• Charge of Proton: +1.602 × 10⁻¹⁹ C.
• Mass of Proton: 1.6726 × 10⁻²⁷ kg/proton.
• Solution:
  • Convert mass to number of protons.
  • Convert protons to coulombs.
  • Calculation: 50 kg / 1.6726 × 10⁻²⁷ kg/proton × 1.602 × 10⁻¹⁹ C/proton = 4.789 × 10⁹ C.
  • Result: Approximately 4.8 Giga coulombs.

Additional Resources

• More difficult problems on electric charge available in a longer video on Coulomb's Law.
• Links to Patreon and YouTube membership for extended versions and worksheets.

Problem 4: Charge Equilibrium of Metal Spheres

• Given: Two metal spheres, one with +20 µC and another with +4 µC.
• Objective: Find new charge after equilibrium.
• Solution:
  • Average the charges: (20 + 4) / 2 = 12 µC each.
  • Result: Both spheres have 12 µC after equilibrium.
• Particle Transfer:
  • Electrons, not protons, are transferred.
  • Charge Transfer: 8 µC from sphere 2 to sphere 1 (negative due to electron transfer).

Calculating Transferred Particles

• Charge Transferred: -8 µC.
• Conversion:
  • Convert to coulombs: -8 µC = -8 × 10⁻⁶ C.
  • Convert to electrons: -8 × 10⁻⁶ C / -1.602 × 10⁻¹⁹ C/electron = 4.99 × 10¹³ electrons.
  • Result: 4.99 × 10¹³ electrons transferred.

Law of Conservation of Electric Charge

• Total electric charge remains constant in a closed system.
• Example: Initial total charge (20 + 4 µC) = Final total charge (12 + 12 µC) = 24 µC.