(a)Two parallel conducting plates in air are connected to theterminals of a 12V

Question

(a)Two parallel conducting plates in air are connected to theterminals of a 12V battery. The capacitance in air is 1.0 F.Then the battery is disconnected and a plexiglass sheet that justfits the gap is inserted between the plates. What is the finalpotential difference in V? What is the stored energy in J? (b) Again the two plates in air are connected to the 12Vbattery. While the battery is still connected, the plexiglass sheetis inserted between the plates. What is the stored energy inJ?
Dielectric constant for plexiglass=3.4
(b) Again the two plates in air are connected to the 12Vbattery. While the battery is still connected, the plexiglass sheetis inserted between the plates. What is the stored energy inJ?
Dielectric constant for plexiglass=3.4

Explanation / Answer

So when increases , C also increases , Initial capacitance = 1 F Final capacitance = 3.4 F
a) When the battery is charged , Total charge stored Q = CV = 12 C As capacitor was disconnected , So charge remains constant, Final C = 3.4F So final voltage = Q/C = 12C / 3.4F = 3.53V Also energy stored =CV2/2 = 3.4*3.532 / 2=22.9 J
b) Here when the battery is connected , Potential remains constant . So V = 12 V Final C = 3.4 F Energy = CV2/2 = 3.4*122/2 = 244.8 J

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