Question

11(a) Define dielectric strength.
(b)(i) An electromagnetic wave has a wavelength shorter than those of radio wave and microwave but longer than that of visible light. Identify the wave. (ii) Name one suitable detector for the wave. (iii) Name one source of the wave.
(c) An oil drop carrying a charge of 1.0×10⁻¹⁹C is found to remain at rest in a uniform electric field of intensity 1,200 NC⁻¹. Calculate the weight of the oil drop.
(d) An RLC series circuit consists of a 100 Ω resistor, 0.05 H inductor and a 25 μF capacitor. A 220 V, 50 Hz mains voltage is applied across the circuit. Calculate the: (i) impedance; (ii) current. (Take π=3.14)

Accepted Answer

(a) Dielectric strength is defined as the maximum voltage that can be applied to a given material without causing it to breakdown. It is the material property that determines the electrical insulation quality of a material. It is measured in volt per metre (Vm⁻¹).

(b)(i) The wave is Infrared wave (its wavelength is shorter than radio and microwaves but longer than visible light).
(ii) Detector: Thermopile / Bolometer / Pneumatic detector / Golay cell / pyroelectric detector
(iii) Source: Human body / Light bulb / Sun

(c) At equilibrium, the electric force balances the weight:
F = W (=mg)
E = F/q → F = E×q = 1200 × 1.0×10⁻¹⁹ = 1.2×10⁻¹⁶N
The weight of the oil drop = 1.2×10⁻¹⁶N

(d) Given R=100Ω, L=0.05H, C=25μF, V=220V, f=50Hz
Impedance Z=√(R²+(XL-XC)²)
XL=2πfL=2π×50×0.05=5π=15.7Ω
XC=1/(2πfC)=1/(2×3.14×50×25×10⁻⁶)=10⁶/7855=127.3Ω
Z=√(100²+(15.7-127.3)²)=√(100²+(-111.6)²)=√(10000+12454.56)=√22454.56=149.85Ω

(ii) Current I=V/Z=220/149.85=1.47A

WAEC Physics 2016 Theory — Question 11

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