Capacitors
Dielectric
RC Circuits
100
With a 4 volt emf, a 4 microfarad capacitor holds 16 micro coulombs of charge. How will this charge change when a new 2 microfarad capacitor is switched in?
The charge on the plates will decrease and become 8 micro coulombs, due to q=CV.
100
Suppose there is a capacitor that is isolated, and a dielectric is added. What decreases and what increases?
Capacitance increases and potential difference decreases (because the amount of charge stays constant).
100
Find the time constant of the loop below:
req = 10 ohms
ceq = 20/3 uf
t = RC = 10 * (20/3 *10 ^-6) = 6.7 * 10^-7
200
Solve for capacitance:
(1/(1/5+1/3.5))+8+(1/(1/1.5+1/(.75+15)))=11.43 microFarads
200
Suppose we have a material of dielectric strength Emax initial and thickness dinitial. We change the thickness so that dnew=2(dinitial). What is the proportion between breakdown potential of the new and initial capacitors?
2
Because breakdown potential (Vmax) = EMaxd, and EMax is the same for both capacitors (they have the same dielectric) the proportion of new/inital = 2.
200
What would be the end behavior of the current on resistor 'R' when the switch is closed towards b and the capacitor C is fully charged initially?
The current through the resistor R will approach 0 as voltage drops comes closer to 0.
300
What is the charge on the plate of the 100 microfarad capacitor and what is the energy across the capacitor?
Answer:
q100=1.39*10^-4Coulombs
U100=9.6605*10^-5Joules
First find total capacitance:
1/(1/20+1/(100+1/(1/60+1/40))) = 17.22microfarads
q across series is the same, so 20 and 40,60,100 trio are the same
qtotal=q20=C20V20=q60,40,100=C60,40,100V60,40,100=CtotalVtotal=17.22*10^-6*10=1.722*10^-4 farads
V20=q20/C20=1.722*10^-4/20*10^-6=8.61V
Voltage is distributed across series capacitors and equal accross parallel capacitors
V60,40,100=V100=10-8.61=1.39V
q100=C100V100=100*10^-6*1.39=1.39*10^-4Coulombs
U100=q^2/2C=(1.39*10^-4)^2/2(100*10^-6)=9.6605*10^-5Joules
300
For this image, given that A = 10*10-2 m2, k1 = 2, k2 = 5, and d = 3*10-13 meters, what is the equivalent capacitance of the capacitor?
Note: use ε0=8.85*10-12 C2/(N*m2)
Ceq = C1+C2 = 2.95 + 7.375 = 10.325 F
300
A simple loop with a capacitor charged to 20 uF and a resistor of 15 kiloohms resistance and a switch is constructed. What will be the equation to find the voltage of the resistor? What will the graph of current vs resistance look like at steady state in a long time?
V(t)r = 20uf (e ^-t/0.3)
RC = 20 uF * 15 kiloohms = 0.3
After a long time in steady-state, the current in the resistor will be close to 0 as there will be no voltage drop.