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ElectrostaticsEasy

Coulomb force between two charges

Two point charges q₁ = +4 μC and q₂ = −6 μC are placed at distance r = 0.30 m in vacuum.\nCalculate: (a) the interaction force, (b) the electric field produced by q₁ at the location of q₂.
Given data
q₁ = 4×10⁻⁶ Cq₂ = −6×10⁻⁶ Cr = 0.30 mk = 8.99×10⁹ N·m²/C²
Review the theory: Elettrostatica
Steps
0 / 2
  1. Calculate the Coulomb force between the two charges (absolute value, in Newton).
  2. Calculate the electric field produced by q₁ at the location of q₂ (in V/m).
Full worked solution
  1. Calculate the Coulomb force between the two charges (absolute value, in Newton).
    F=k∣q1∣∣q2∣r2=8.99×109⋅4×10−6⋅6×10−6(0.30)2F = k \dfrac{|q_1||q_2|}{r^2} = 8.99\times10^9 \cdot \dfrac{4\times10^{-6} \cdot 6\times10^{-6}}{(0.30)^2}
    F=k∣q1∣∣q2∣r2=8.99×109×4×10−6×6×10−6(0.30)2=8.99×109×24×10−120.09=8.99×109×2.667×10−10≈2.40 NF = k\frac{|q_1||q_2|}{r^2} = 8.99\times10^9 \times \frac{4\times10^{-6} \times 6\times10^{-6}}{(0.30)^2} = 8.99\times10^9 \times \frac{24\times10^{-12}}{0.09} = 8.99\times10^9 \times 2.667\times10^{-10} \approx \mathbf{2.40\,N}. Attractive force since the charges have opposite signs.
  2. Calculate the electric field produced by q₁ at the location of q₂ (in V/m).
    E1=k∣q1∣r2=8.99×109⋅4×10−6(0.30)2E_1 = k \dfrac{|q_1|}{r^2} = 8.99\times10^9 \cdot \dfrac{4\times10^{-6}}{(0.30)^2}
    E1=k∣q1∣r2=8.99×109×4×10−60.09=8.99×109×4.444×10−5≈3.996×105 V/m≈400 kV/mE_1 = k\frac{|q_1|}{r^2} = 8.99\times10^9 \times \frac{4\times10^{-6}}{0.09} = 8.99\times10^9 \times 4.444\times10^{-5} \approx \mathbf{3.996\times10^5\,V/m} \approx \mathbf{400\,kV/m}. Verification: F=∣q2∣ E1=6×10−6×4×105=2.40 NF = |q_2|\,E_1 = 6\times10^{-6} \times 4\times10^5 = 2.40\,\mathrm{N} ✓.
Result:F = 2.40 N (attractive). E₁ = 400 kV/m at the location of q₂.
TheoryElettrostaticaFormula sheetPhysics 2

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