This physics problem tests your understanding of fundamental physical laws and their applications. The step-by-step solution below breaks down the problem using relevant equations and physical reasoning.
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23. (a) Define capacitance. Capacitance is the measure of a capacitor's ability to store an electric charge. It is defined as the ratio of the magnitude of the charge stored on either conductor to the potential difference between the conductors. (b) Two plates of a parallel-plate capacitor are 0.8mm apart and each has an area 4cm^2. Given that the potential difference between the plates is 100V, (i) Find the capacitance of the capacitor (take _0 = 8.85 × 10^-12 FM^-1). Step 1: Convert given values to SI units. Distance between plates, d = 0.8 mm = 0.8 × 10^-3 m Area of plates, A = 4 cm^2 = 4 × (10^-2 m)^2 = 4 × 10^-4 m^2 Permittivity of free space, _0 = 8.85 × 10^-12 F/m Step 2: Use the formula for the capacitance of a parallel-plate capacitor. C = (_0 A)/(d) Step 3: Substitute the values and calculate C. C = (8.85 × 10^-12 F/m) × (4 × 10^-4 m^2)0.8 × 10^-3 m C = 35.4 × 10^-160.8 × 10^-3 F C = 44.25 × 10^-13 F C = 4.425 × 10^-12 F The capacitance of the capacitor is 4.425 × 10^-12 F. (ii) Calculate the charge stored in the capacitor. Step 1: Use the formula relating charge, capacitance, and potential difference. Q = CV Given potential difference, V = 100 V Capacitance, C = 4.425 × 10^-12 F (from part b(i)) Step 2: Substitute the values and calculate Q. Q = (4.425 × 10^-12 F) × (100 V) Q = 442.5 × 10^-12 C Q = 4.425 × 10^-10 C The charge stored in the capacitor is 4.425 × 10^-10 C. (c) State two applications of a capacitor. • Capacitors are used for energy storage, such as in camera flashes and defibrillators, where they release a large amount of energy quickly. • They are used in filtering circuits to smooth out voltage fluctuations in power supplies or to block DC current while allowing AC current to pass. Send me the next one 📸