1. The plates of a parallel-plate capacitor are 5 mm apart. The area of each pla
ID: 2245845 • Letter: 1
Question
1. The plates of a parallel-plate capacitor are 5 mm apart. The area of each plate is 102x10^-4 m^2. A potential difference of Vo= 75V is applied between the plates. The plates are filled with a dielectric with constant of k=2.7
a) What is the capacitance C0 before the dielectric slab is inserted?
b) What free charge appears on the plates?
c) What is the capacitance after inserting the dielectric?
d) What is the potential difference V between the plates after the slab has been introduced?
2. A 120-V fish tank heater has a power rating of 112 W.
a) Calculate the current through the heater
b) Calculate the resistance of the heater
c) Find the total electric energy transformed if the heater was turned on for half hour.
Explanation / Answer
As we know that
C = epsilonA/d
= (8.854*10^-12*102*10^-4)/0.005
= 1.806*10^-11 F
Q = CV
= (1.806*10^-11)*(75)
= 1.3545*10^-9 C
After Inserting
Capactince = kappa*Initial apictance
= 2.7*(1.806*10^-11)
= 4.876*10^-11 F
As Q remains Same
Therefore
Q = CV
V = (1.3545*10^-9)/(4.876*10^-11)
= 27.779 V
Power = VI
I = P/V
= (112/120)
= 0.9333 A
V = IR
Therefore
R = (120/0.933)
= 128.57 ohm
Energy Lost = Power*time
= 112*0.5*60*60
= 201600 J
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