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1a) Coulomb's Law states that:

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1a) Coulomb's Law states that:

  • The magnitude of the electrostatic 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. It is repulsive if the charges have the same sign and attractive if they have opposite signs.

1b)

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

    Charge 1, q1=2μC=2×10−6Cq_1 = 2\mu C = 2 \times 10^{-6} C
    Charge 2, q2=12μC=12×10−6Cq_2 = 12\mu C = 12 \times 10^{-6} C
    Distance, r=650mm=0.650mr = 650mm = 0.650 m
    Coulomb's constant, k=14πϵ0≈8.9875×109 N⋅m2/C2k = \frac{1}{4\pi\epsilon_0} \approx 8.9875 \times 10^9 \, \text{N} \cdot \text{m}^2/\text{C}^2

  2. Apply Coulomb's Law formula for the magnitude of the force.

    F=k∣q1q2∣r2F = k \frac{|q_1 q_2|}{r^2}

  3. Substitute the values into the formula.

    F=(8.9875×109 N⋅m2/C2)∣(2×10−6 C)(12×10−6 C)∣(0.650 m)2F = (8.9875 \times 10^9 \, \text{N} \cdot \text{m}^2/\text{C}^2) \frac{|(2 \times 10^{-6} \, \text{C})(12 \times 10^{-6} \, \text{C})|}{(0.650 \, \text{m})^2}

  4. Calculate the product of the charges and the square of the distance.

    ∣q1q2∣=(2×10−6)(12×10−6)=24×10−12 C2|q_1 q_2| = (2 \times 10^{-6})(12 \times 10^{-6}) = 24 \times 10^{-12} \, \text{C}^2 r2=(0.650)2=0.4225 m2r^2 = (0.650)^2 = 0.4225 \, \text{m}^2

  5. Perform the final calculation.

    F=(8.9875×109)24×10−120.4225F = (8.9875 \times 10^9) \frac{24 \times 10^{-12}}{0.4225} F=215.7×10−30.4225F = \frac{215.7 \times 10^{-3}}{0.4225} F≈0.5105 NF \approx 0.5105 \, \text{N}

    The magnitude of the force between the charges is approximately 0.511 N.

    0.511 N\boxed{0.511 \, \text{N}}
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