1. A 6.5-cm by 2.3-cm parallel plate capacitor has the plates separated by a dis

Question

1. A 6.5-cm by 2.3-cm parallel plate capacitor has the plates separated by a distance of 2.1 mm.

(a) When 4.0 10-11 C of charge is placed on this capacitor, what is the electric field between the plates?
in V/m

(b) If a dielectric with dielectric constant of 4.8 is placed between the plates while the charge on the capacitor stays the same, what is the electric field in the dielectric?
in V/m

2. A parallel plate capacitor has a charge of 0.040 µC on each plate with a potential difference of 250 V. The parallel plates are separated by 0.40 mm of air. What energy is stored in this capacitor?

Explanation / Answer

The area of the plates = 6.5 cm * 2.3 cm = 14.95 cm^2 = 14.95 * 10^-4 m^2 The separation between the plates, d = 2.1 mm = 2.1 * 10^-3 m The charge on the plates, q = 4 * 10^-11 C The capacity of parallel plate capacitor, C = 0A/d C = ( 8.85 * 10^-12 * 14.95 * 10^-4 ) / ( 2.1 * 10^-3 ) C = 6.2 * 10^-12 F a) We have a formula, v = q/c and also v = Ed From these we can write that, Ed = q/c So, E = q/dc = ( 4 * 10^-11 C ) / (2.1 * 10^-3 m )(6.2 * 10^-12 F) E = 0.31 * 10^4 = 3100 V/m b) When a dielectric material of constant k will place between the plates its capacity will   increases 'k' times of its original So new capacitance, c' = kc = 4.8 * (6.2 * 10^-12 F) =  29.8 * 10^-12 F 2. The energy stored in the capacitor is (1/2)qv. 2. The energy stored in the capacitor is (1/2)qv.

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