(a) Explain briefly dielectric strength.
(b) An electromagnetic wave has its wavelength shorter than those of radiowave and microwave but longer than that of visible light.
(i) 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 x 10\(^{-19 }\)C is found to remain at rest in a uniform electric field of intensity 1200 NC\(^{-1}\). Calculate the weight of the oil drop.
(d) An RLC series circuit consists of a 100\(\Omega\) resistor, 0.05 H inductor and a 25 \(\mu\) capacitor. A 220 V, 50 Hz mains voltage is applied across the circuit. Calculate the:
(i) impedance;
(ii) current. (\(\pi\) = 3.14)
(a) Dielectric strength: the maximum electric field (or maximum p.d. per unit thickness) that an insulator (dielectric) can withstand without breaking down and allowing charge to conduct through it.
(b) The wave is infrared radiation.
- (i) Infrared (heat) radiation.
- (ii) Detector: a thermopile / bolometer / blackened thermometer (a special infrared-sensitive photodiode is also acceptable).
- (iii) Source: any hot body, e.g. the Sun, an electric heating element or a warm object.
(c) Weight of the oil drop. Since the drop is at rest, the electric force balances the weight:
\[ W = qE = (1.0\times10^{-19})(1200) = 1.2\times10^{-16}\,\text{N} \]
(d) RLC series circuit. \(R=100\,\Omega\), \(L=0.05\,\text{H}\), \(C=25\,\mu\text{F}=25\times10^{-6}\,\text{F}\), \(f=50\,\text{Hz}\).
\[ X_L = 2\pi f L = 2(3.14)(50)(0.05)=15.7\,\Omega \]\[ X_C = \frac{1}{2\pi f C}=\frac{1}{2(3.14)(50)(25\times10^{-6})}=\frac{1}{7.85\times10^{-3}}=127.4\,\Omega \]
(i) Impedance:
\[ Z=\sqrt{R^2+(X_C-X_L)^2}=\sqrt{100^2+(127.4-15.7)^2}=\sqrt{10000+12477}=\sqrt{22477}=150\,\Omega \]
(ii) Current:
\[ I=\frac{V}{Z}=\frac{220}{150}=1.47\,\text{A} \]
(a) Dielectric strength: the maximum electric field (or maximum p.d. per unit thickness) that an insulator (dielectric) can withstand without breaking down and allowing charge to conduct through it.
(b) The wave is infrared radiation.
- (i) Infrared (heat) radiation.
- (ii) Detector: a thermopile / bolometer / blackened thermometer (a special infrared-sensitive photodiode is also acceptable).
- (iii) Source: any hot body, e.g. the Sun, an electric heating element or a warm object.
(c) Weight of the oil drop. Since the drop is at rest, the electric force balances the weight:
\[ W = qE = (1.0\times10^{-19})(1200) = 1.2\times10^{-16}\,\text{N} \]
(d) RLC series circuit. \(R=100\,\Omega\), \(L=0.05\,\text{H}\), \(C=25\,\mu\text{F}=25\times10^{-6}\,\text{F}\), \(f=50\,\text{Hz}\).
\[ X_L = 2\pi f L = 2(3.14)(50)(0.05)=15.7\,\Omega \]\[ X_C = \frac{1}{2\pi f C}=\frac{1}{2(3.14)(50)(25\times10^{-6})}=\frac{1}{7.85\times10^{-3}}=127.4\,\Omega \]
(i) Impedance:
\[ Z=\sqrt{R^2+(X_C-X_L)^2}=\sqrt{100^2+(127.4-15.7)^2}=\sqrt{10000+12477}=\sqrt{22477}=150\,\Omega \]
(ii) Current:
\[ I=\frac{V}{Z}=\frac{220}{150}=1.47\,\text{A} \]
