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9702 · 18.3

Electric force between point charges — practice questions

Practice and worked examples for 9702 Electric force between point charges. Short previews only — attempt the full question in MarkScheme against the official scheme.

Worked example 1

Two point charges, Q1=+4.0×10−6 CQ_1 = +4.0 \times 10^{-6} \text{ C} and Q2=−3.0×10−6 CQ_2 = -3.0 \times 10^{-6} \text{ C}, are separated by a distance of $0.20 \text{ m}$ in a vacuum. Calculate the magnitude of the electrostatic force between them and state its nature.

Show solution outline
  1. Identify knowns and state the formula:
    • Q1=+4.0×10−6 CQ_1 = +4.0 \times 10^{-6} \text{ C}
    • Q2=−3.0×10−6 CQ_2 = -3.0 \times 10^{-6} \text{ C}
    • r=0.20 mr = 0.20 \text{ m}
    • Coulomb's constant, ke=8.99×109 N m2 C−2k_e = 8.99 \times 10^9 \text{ N m}^2 \text{ C}^{-2}
    • Formula: F=ke∣Q1Q2∣r2F = k_e \frac{|Q_1 Q_2|}{r^2}
  2. Substitute values into the formula: We use the magnitudes of the charges for the calculation. F=(8.99×109)(4.0×10−6)(3.0×10−6)(0.20)2F = (8.99 \times 10^9) \frac{(4.0 \times 10^{-6})(3.0 \times 10^{-6})}{(0.20)^2}
  3. Calculate the force magnitude: F=(8.99×109)1.2×10−110.040F = (8.99 \times 10^9) \frac{1.2 \times 10^{-11}}{0.040} F=2.697 NF = 2.697 \text{ N}
  4. Determine the nature and state the final answer: Since the charges have opposite signs, the force is attractive. Rounding to two significant figures (consistent with the input values), the magnitude of the force is 2.7 N.

Worked example 2

Three point charges are placed along the x-axis. Charge Q1=+8.0 nCQ_1 = +8.0 \text{ nC} is at the origin (x=0x=0), charge Q2=−3.0 nCQ_2 = -3.0 \text{ nC} is at x=4.0 cmx = 4.0 \text{ cm}, and charge Q3=+5.0 nCQ_3 = +5.0 \text{ nC} is at x=6.0 cmx = 6.0 \text{ cm}. Calculate the net electrostatic force on charge Q3Q_3.

Show solution outline
  1. Apply the Principle of Superposition: The net force on Q3Q_3 is the vector sum of the force from Q1Q_1 (F13F_{13}) and the force from Q2Q_2 (F23F_{23}). Let's define the positive direction as to the right (along the +x axis).
  2. Calculate the force from Q1Q_1 on Q3Q_3 (F13F_{13}):
    • Charges Q1Q_1 and Q3Q_3 are both positive, so the force is repulsive. F13F_{13} is directed to the right (+ve).
    • Distance: r13=6.0 cm=0.060 mr_{13} = 6.0 \text{ cm} = 0.060 \text{ m}.
    • Magnitude: F13=ke∣Q1Q3∣r132=(8.99×109)(8.0×10−9)(5.0×10−9)(0.060)2F_{13} = k_e \frac{|Q_1 Q_3|}{r_{13}^2} = (8.99 \times 10^9) \frac{(8.0 \times 10^{-9})(5.0 \times 10^{-9})}{(0.060)^2}
    • F13=(8.99×109)4.0×10−173.6×10−3=9.989×10−5 NF_{13} = (8.99 \times 10^9) \frac{4.0 \times 10^{-17}}{3.6 \times 10^{-3}} = 9.989 \times 10^{-5} \text{ N}. So, F13=+1.0×10−4 NF_{13} = +1.0 \times 10^{-4} \text{ N} (to 2 s.f.).
  3. Calculate the force from Q2Q_2 on Q3Q_3 (F23F_{23}):
    • Charges Q2Q_2 and Q3Q_3 are opposite, so the force is attractive. F23F_{23} is directed to the left (-ve).
    • Distance: r23=6.0 cm−4.0 cm=2.0 cm=0.020 mr_{23} = 6.0 \text{ cm} - 4.0 \text{ cm} = 2.0 \text{ cm} = 0.020 \text{ m}.
    • Magnitude: F23=ke∣Q2Q3∣r232=(8.99×109)(3.0×10−9)(5.0×10−9)(0.020)2F_{23} = k_e \frac{|Q_2 Q_3|}{r_{23}^2} = (8.99 \times 10^9) \frac{(3.0 \times 10^{-9})(5.0 \times 10^{-9})}{(0.020)^2}
    • F23=(8.99×109)1.5×10−174.0×10−4=3.371×10−4 NF_{23} = (8.99 \times 10^9) \frac{1.5 \times 10^{-17}}{4.0 \times 10^{-4}} = 3.371 \times 10^{-4} \text{ N}. The direction is negative, so we use −3.37×10−4 N-3.37 \times 10^{-4} \text{ N}.
  4. Calculate the net force (FnetF_{net}):
    • Fnet=F13+F23F_{net} = F_{13} + F_{23} (vector sum)
    • Fnet=(+1.00×10−4 N)+(−3.37×10−4 N)F_{net} = (+1.00 \times 10^{-4} \text{ N}) + (-3.37 \times 10^{-4} \text{ N})
    • Fnet=−2.37×10−4 NF_{net} = -2.37 \times 10^{-4} \text{ N}
  5. Final Answer: The net electrostatic force on charge Q3Q_3 is $2.4 \times 10^{-4} \text{ N}$ (to 2 s.f.) directed to the left (in the negative x-direction).
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