A parallel plate capacitor has a plate separation of 4.80 mm and a plate area of

Question

A parallel plate capacitor has a plate separation of 4.80 mm and a plate area of 25.0 cm^2. Initially there is only air between the plates. The plates are charged to a potential difference of 120 V and then disconnected from the source. a) What is the capacitance of the capacitor? b) What is the charge on the capacitor? c) How much energy is stored by the capacitor? The charged capacitor is now immersed in distilled water (which is assumed to be an insulator). d) What is the capacitance after immersion in the water? e) What is the charge on the capacitor after immersion in the water? f) What is the potential difference between the capacitor plates after immersion in the water? g) What is the energy stored by the capacitor after immersion in the water?

Explanation / Answer

d = 4.8 mm , A = 25 cm^2, V = 120 V

a) C = Aeo/d

C = 25*10^-4*8.85*10^-12/(4.8*10^-3)

C = 4.6*10^-12 F

b) Q = VC

Q = 120*4.6*10^-12 = 5.52*10^-10 C

c) U = (1/2)CV^2

U = 0.5*4.6*10^-12*120*120 = 3.3*10^-8 J

d) dielectric constant of distilled water k =80

C = kco = 80*4.6*10^-12 = 368*10^-12 F

e) Q = Qo = 5.52*10^-10 C

f) V = Vo/k = 120/80 = 1.5 V

g) U = Uo/k = 3.3*10^-8/80 = 4.125*10^-10 J

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