What is the capacitance, C = ? of the following parallel plate capacitor, where
ID: 1969677 • Letter: W
Question
What is the capacitance, C = ? of the following parallel plate capacitor, where the plate separation is, d = 5 mm, and radius of places is R = 50 mm, and glass is between the plates Note, Area of plates = are SI units. Assume that there is no net charge (q-0) initially on the parallel plate capacitor. At time t = 0, the switch (SWI) is closed and the parallel plate capacitor begins to charge. When completely charged, what is the maximum charge q on the capacitor? What is work, in Joules [J], done by the battery to change the capacitor to a maximum charge q ? Hint, this is the same as energy stored in capacitor. Which plate is positively charged, 1 or 2? Circle choice here.Explanation / Answer
SOLUTION: (a) Plate speration,d=5 mm =(5 mm)(10-3 m/ 1mm) =0.005 m Radius of the plater, r=50 mm =(50 mm)(10-3 m/ 1mm) = 0.05 m Area of the plates, A=r2 =(3.14)(0.05 m)2 =0.00785 m2 Di-electric constant, k=4.5 Voltage of the battery, V=5 V resistence of the circuit, R=20 k Capacitence of the parallel plate capacitor, C=Ao/d =(0.00785 m2 )(8.854 x 10-12 Nm2/C2)/(0.005 m) =13.9 x10-12 F =13.9 pF ___________________________________________________ ___________________________________________________ (b) Maximum charge on the capacitor, qmax=CV =(13.9x10-12 F)(5 V) =69.5 x10-12 C =69.5 pC _________________________________________________ _________________________________________________ (c) W=Vqmax =(5 V)(69.5 x 10-12 C) =347.5 x 10-12 J _________________________________________________ _________________________________________________ (d) First one, choice :(1) =(5 V)(69.5 x 10-12 C) =347.5 x 10-12 J _________________________________________________ _________________________________________________ (d) First one, choice :(1)Related Questions
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