This set of problems is designed to help you prepare for the lab on capacitors.

Question

This set of problems is designed to help you prepare for the lab on capacitors. Here are some things you should know.
The charge on a capacitor is proportional to the voltage. The constant of proportionality is the capacitance:
q=VC.
The simplest type of capacitor is a parallel plate capacitor consisting of twoparallel conducting plates...DUH. If the plates each have area A and
are separated by a distance d, the capacitance will be
C=Ke0A/d,
where e0 is the permittivity of free space (you measured that last week) and K is the dielectric constant of what ever material is between the plates. If the space between the plates is filled with vacuum, the dielectric constant is exactly 1. Air is pretty darn close to 1.
When a capacitor is charged in a series circuit with a resistor, the charge builds up with time. Consider the circuit below.



When the switch, S, is thrown to position A, the battery, with voltage V0, will supply a current through the resistor, R, that charges the capacitor C. The voltage across the capacitor at any given time is given by
V=V0(1-e-t/RC).
The product RC is known as the time constant. After waiting a time equal to 1 time constant, the voltage on the capacitor is (1-e-1)=0.632 of V0. After 5RC the capacitor is nearly fully charged and has a voltage of 0.993V0. Eventually, the voltage builds up to V0.
When the switch is thrown to position B, the capacitor will discharge through the resistor. The voltage as a fuction of time will be given by
V=V0(e-t/RC).
After a long enough time, the capacitor will fully discharge the resistor.

1. [1pt]
A parallel plate capacitor has plates with length and width of 5.7 cm and are separated by 7.0 mm. The capacitance is measured to be 1.479×10-11 F. What is the dielectric constant of the material between the plates?

Correct, computer gets: 3.60

2. [1pt]
Consider a series circuit like the one shown above with a resistor of R=80. kilo-ohms and a capacitor of C=8.0 uF. What is the time constant of the circuit?

Correct, computer gets: 0.64 s


Hint: Don't forget to convert to Farads and ohms from uF and kilo-ohms.
3. [1pt]
If the battery has a voltage of 12 V, what is the voltage across the capacitor after the switch has been in position A for 2 time constants?


Correct, computer gets: 10.38 V
4. [1pt]
What is the final charge on the capacitor after a long time?


Correct, computer gets: 9.60E-05 C
5. [1pt]
Now let's say that the capacitor has been fully charged and we throw the switch to position B. How long will it take for the voltage on the capacitor to be 0.891 V?


6. [1pt] What is the charge on the capacitor at this time?

Answer: ____________

Explanation / Answer

1. Area A = 5.7 cm * 5.7 cm=32.49 cm ^ 2 = 32.49 * 10 ^ -4 m^2 separation d = 7.0 mm= 7.0 * 10 ^ -3 m . The capacitance C = 1.479×10^-11F we know C = keA / d from this the dielectric constant of the materialbetween the plates k = C d / eA where e = permitivity of free space = 8.85 * 10 ^-12 C ^ 2/ Nm^2 plug the values weget k = 3.60 2. resistance R =80. kilo-ohms= 80 * 10 ^ 3 ohm capacitance C=8.0 uF= 8.0 * 10 ^ -6F the time constant of the circuit = RC = 0.64 s 3. voltage V = 12 V the voltage across the capacitor after the switch hasbeen in position A for 2 time constants is V ' = V[1-e^( -t/RC)] where t = 2RC V ' = V [ 1- e^ -2 ] = 10.3759 Volt 4. the final charge on the capacitor after a longtime Q = CV where C = 8.0*10^-6 F V = 12Volt plug the values we get Q = 76.0 * 10 ^ -6C 5. the voltage on the capacitor after time t is V " = 0.86465V we know V " = V e -t/RC -t/RC = ln ( V " / V ) = -0.1454

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