1.) Consider a parallel plate capacitor with plates of area1.04 m 2 whose plates

Question

1.) Consider a parallel plate capacitor with plates of area1.04 m2whose plates are 1.30 cm apart. Thegap between the plates is filled with air (assume thatair is unity) and thepositive plate has a charge of 11.4 nC on it while the negative plate has anegative charge of equal magnitude on it.

The capacitor has its plate-spacing reduced to 1.01 mm, and the plates are connected to a1.55 V battery.

The capacitor is disconnected from the battery and the platesare returned to their initial spacing.

The space between the plates is now filled with a fluid with adielectric constant of 4.43.

The area of overlap between the capacitor plates is increased to2.31 m2.

Explanation / Answer

. 01.   C = *A/d 02.   C = Q/V . 03.   C1 = (8.854E-12 Farads/meter)*(1.04m2)/(13.0E-3 m) = 708.335E-12 Farads 04.   V = Q/C = (11.4E-9 Coulombs)/(708.32E-12Farads) = 16.094volts . 05.   C2 = (8.854E-12 Farads/meter)*(1.04m2)/(1.01E-3 m) = 9.117E-9 Farads 06.   Q = C*V = (9.12E-9 Farads)*(1.55 volts) =14.13E-9 Coulombs = 14.13nC . 07.   C3 = (8.854E-12 Farads/meter)*(1.04m2)/(13.0E-3 m) = 708.335E-12 Farads 08.   V = Q/C = (14.13E-9 Coulombs)/(708.32E-12Farads) = 19.948volts . 09.   C4 = (r)*C3 =(4.43)*(708.32E-12 Farads) = 3.138E-9 Farads 10.    V = Q/C = (14.13E-9 Coulombs)/(3.138E-9Farads) = 4.493volts . 11.   C5 = (4.43)*(8.854E-12 Farads/meter)*(2.31m2)/(13.0E-3 m) = 6.970E-9 Farads 12.   Q = 14.1E-9Coulombs      ;   doesnot change from '10' . 13.   V = Q/C = (14.1E-9 Coulombs)/(6.97E-9 Farads)= 2.023 volts .

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