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A parallel-plate capacitor has an area of 4.59 cm2, and the plates are separated by 1.28 mm with air between them. it stores a charge of 334 pc. what is the potential difference across the plates of the capacitor?

Answer :

tatamaraweeks74
 a) 
Capacitance = k x ε° x area / separation 
ε° = 8.854 10^-12 F/ m 
k = 2.4max 
average k = 0.78 / 1.27 * 2.4 +(1.27- 0.78) / 1.27 * 1 = 1.474 + 0.386 = 1.86 
(61.4 % separation k = 2.4 --- 38.6 % k = 1 air --- average k = 0.614 * 2.34 + 0.386 * 1 = 1.86 
area = 145 cm2 = 0.0145 m2 
separation = 1.27 cm 0.0127 m 

C = 1.86 * 8.854 10^-12 * 0.0145 / 0.0127 = 18.8 pF 
b) Q = C * V --- 18.8 * 83 = 1560.4 pC = 1.5604 nC 
c) E = V / d = 83 / 0.0127 = 6535.4 V/m 

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