A parallel-plate capacitor with plate area A = 1.5 m^2 and plane separation d =

Question

A parallel-plate capacitor with plate area A = 1.5 m^2 and plane separation d = 45mm is connected to 50 V battery. a) Determine the charge on the capacitor, the electric field, the capacitance, and the energy stored in the capacitor. b) with the capacitor still connected to the battery, slab of plastic with dielectric strength K = 3 is placed between the plates of the capacitor, so that the gap in completely filled with the dielectric When are the new values of charge, electric field, capacitance, and the energy U stored in the capacitor?

Explanation / Answer

A) Capacitance C = Aeo/d

= 1.5*8.85e-12/0.0045

= 2.95*10^-9 F

Charge Q = CV = 2.95e-9*50

= 1.475*10^-7 C

Electric field E = V/d = 50/0.0045 = 11111 V/m

Energy stored = 0.5CV^2 = 0.5*2.95e-9*50^2

= 3.69*10^-6 J

B) capacitance C = kCo

= 3*2.95e-9

= 8.85*10^-9 F

Charge Q = CV = 8.85e-9*50

= 4.425*10^-7 C

Electric field E = V/d = 11111 N/C

Energy E = 0.5CV^2

= 0.5*8.85e-9*50^2

= 1.106*10^-5 J

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