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4a. Define the term resonance as applied to ac circuits. Sketch a resonance curve as applied to RC circuits and hence deduce the equation for resonant frequency.

Physics

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1a. State the laws of electrostatics.

The fundamental laws of electrostatics are:

  • Coulomb's Law: This law states that the force between two point charges is directly proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance between them. The force acts along the line joining the two charges. Mathematically, the magnitude of the force FF between two charges q1q_1 and q2q_2 separated by a distance rr is given by: F=k∣q1q2∣r2F = k \frac{|q_1 q_2|}{r^2} where kk is Coulomb's constant (k≈8.987×109 N⋅m2/C2k \approx 8.987 \times 10^9 \text{ N}\cdot\text{m}^2/\text{C}^2).
  • Principle of Superposition: This principle states that the total electrostatic force on a given charge due to a collection of other charges is the vector sum of the individual forces exerted on that charge by each of the other charges, acting independently.

1b. Two electric charges of +15μC+15\mu C and −10μC-10\mu C are placed 20cm apart in air. Calculate the attraction between the charges.

  1. Identify the given values and convert to SI units.

    • Charge 1: q1=+15μC=+15×10−6 Cq_1 = +15\mu C = +15 \times 10^{-6}\text{ C}
    • Charge 2: q2=−10μC=−10×10−6 Cq_2 = -10\mu C = -10 \times 10^{-6}\text{ C}
    • Distance: r=20cm=0.20 mr = 20\text{cm} = 0.20\text{ m}
    • Coulomb's constant in air (approximately vacuum): k=8.987×109 N⋅m2/C2k = 8.987 \times 10^9\text{ N}\cdot\text{m}^2/\text{C}^2
  2. Apply Coulomb's Law.

    The magnitude of the force FF between the two charges is given by:

    F=k∣q1q2∣r2F = k \frac{|q_1 q_2|}{r^2}
  3. Substitute the values and calculate the force.

    F=(8.987×109 N⋅m2/C2)∣(+15×10−6 C)×(−10×10−6 C)∣(0.20 m)2F = (8.987 \times 10^9\text{ N}\cdot\text{m}^2/\text{C}^2) \frac{|(+15 \times 10^{-6}\text{ C}) \times (-10 \times 10^{-6}\text{ C})|}{(0.20\text{ m})^2} F=(8.987×109)∣−150×10−12∣(0.04) NF = (8.987 \times 10^9) \frac{|-150 \times 10^{-12}|}{(0.04)}\text{ N} F=(8.987×109)150×10−120.04 NF = (8.987 \times 10^9) \frac{150 \times 10^{-12}}{0.04}\text{ N} F=(8.987×109)×(3750×10−12) NF = (8.987 \times 10^9) \times (3750 \times 10^{-12})\text{ N} F=8.987×3.75 NF = 8.987 \times 3.75\text{ N} F≈33.70 NF \approx \boxed{33.70\text{ N}}

    Since the charges have opposite signs (+15μC+15\mu C and −10μC-10\mu C), the force between them is attractive.

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