Question

9a(i) State coulomb's law of electrostatics.
(ii) The electron and proton of a hydrogen atom are separated by a mean distance of 5.2×10⁻¹¹m. Calculate the magnitude of the electrostatic force between the particles. [e=1.6×10⁻¹⁹C, (4πε₀)⁻¹=9.0×10⁹mF⁻¹]
(b)(i) The diagram below shows a potential divider circuit. Show that Vₒᵤₜ=Vᵢₙ(R₂/(R₁+R₂)).
(ii) If Vᵢₙ/Vₒᵤₜ=2.5 and R₁=30Ω, calculate R₂.

Accepted Answer

(a)(i) Coulomb's law of electrostatics states that the electrostatic force between two charged bodies is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.

(ii) Given: r=5.2×10⁻¹¹m, e=q=1.6×10⁻¹⁹C, k=(4πε₀)⁻¹=9×10⁹mF⁻¹
Using F=kq₁q₂/r²:
F = (9×10⁹×1.6×10⁻¹⁹×1.6×10⁻¹⁹)/(5.2×10⁻¹¹)²
F = 23.04×10⁻²⁹/27.04×10⁻²²
F = 8.52×10⁻⁸N.

(b)(i) Let the current through the circuit be I.
Vᵢₙ=I(R₁+R₂) ...(1)
Vₒᵤₜ=IR₂ ...(2)
Dividing (2) by (1): Vₒᵤₜ/Vᵢₙ = R₂/(R₁+R₂)
Therefore, Vₒᵤₜ=Vᵢₙ(R₂/(R₁+R₂)).

(ii) Given Vᵢₙ/Vₒᵤₜ=2.5, R₁=30Ω. Inverting the equation: Vᵢₙ/Vₒᵤₜ=(R₁+R₂)/R₂ ⟹ 2.5=(30+R₂)/R₂ ⟹ 2.5R₂=30+R₂ ⟹ 1.5R₂=30 ⟹ R₂=20Ω.

WAEC Physics 2022 Theory — Question 9

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