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1.Consider the build-up of the charge on a capacitor as described in the introducation to this lab. Assume that you start with an initially uncharged capacitor, as in figure 1 and then you close the switch to a battery of Voltage V0. After that time, how many time constants must elapse before the charge reaches 39%, 63%, and 86% of its final value?

2.For the same case you were considering in problem 1, asume that the capacitor is now fully charged. Now the switch is flipped to position 2, and the charge on the capacitor begins to decay away. After that time, how many time constants much elapse before the charge reaches 61%, 37% and 13% of its initial value (the value right before you flipped the switch to position 2)?

 

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Sadha
SadhaLv10
26 Mar 2021

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Related textbook solutions

Physics: Principles with Applications
7th Edition, 2013
Giancoli
ISBN: 9780321625922

Related questions

This set of problems is designed to help you prepare for the lab oncapacitors. Here are some things you should know.
The charge on a capacitor is proportional to the voltage. Theconstant of proportionality is the capacitance:
q=VC.
The simplest type of capacitor is a parallel plate capacitorconsisting of twoparallel conducting plates...DUH. If the plateseach 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 lastweek) and K is the dielectric constant of what ever material isbetween the plates. If the space between the plates is filled withvacuum, the dielectric constant is exactly 1. Air is pretty darnclose 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, withvoltage V0, will supply a current through the resistor, R, thatcharges the capacitor C. The voltage across the capacitor at anygiven time is given by
V=V0(1-e-t/RC).
The product RC is known as the time constant. After waiting a timeequal to 1 time constant, the voltage on the capacitor is(1-e-1)=0.632 of V0. After 5RC the capacitor is nearly fullycharged and has a voltage of 0.993V0. Eventually, the voltagebuilds up to V0.
When the switch is thrown to position B, the capacitor willdischarge through the resistor. The voltage as a fuction of timewill be given by
V=V0(e-t/RC).
After a long enough time, the capacitor will fully discharge theresistor.

1. [1pt]
A parallel plate capacitor has plates with length and width of 5.7cm and are separated by 7.0 mm. The capacitance is measured to be1.479×10-11 F. What is the dielectric constant of the materialbetween the plates?

Correct, computer gets: 3.60

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

Correct, computer gets: 0.64 s


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


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 wethrow the switch to position B. How long will it take for thevoltage on the capacitor to be 0.891 V?


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

Answer: ____________

This set of problems is designed to help you prepare for the lab oncapacitors. Here are some things you should know.
The charge on a capacitor is proportional to the voltage. Theconstant of proportionality is the capacitance:
q=VC.
The simplest type of capacitor is a parallel plate capacitorconsisting of twoparallel conducting plates...DUH. If the plateseach 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 lastweek) and K is the dielectric constant of what ever material isbetween the plates. If the space between the plates is filled withvacuum, the dielectric constant is exactly 1. Air is pretty darnclose 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, withvoltage V0, will supply a current through the resistor, R, thatcharges the capacitor C. The voltage across the capacitor at anygiven time is given by
V=V0(1-e-t/RC).
The product RC is known as the time constant. After waiting a timeequal to 1 time constant, the voltage on the capacitor is(1-e-1)=0.632 of V0. After 5RC the capacitor is nearly fullycharged and has a voltage of 0.993V0. Eventually, the voltagebuilds up to V0.
When the switch is thrown to position B, the capacitor willdischarge through the resistor. The voltage as a fuction of timewill be given by
V=V0(e-t/RC).
After a long enough time, the capacitor will fully discharge theresistor.

--------------------------------------------------------------------------------


For multiple choice questions, enter the letter corresponding tothe correct answer.

--------------------------------------------------------------------------------

1. [1pt]
A parallel plate capacitor has plates with length and width of 5.7cm and are separated by 7.0 mm. The capacitance is measured to be1.479×10-11 F. What is the dielectric constant of the materialbetween the plates?

Correct, computer gets: 3.60


Hint: Don't forget to convert distances to meters.



--------------------------------------------------------------------------------

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

Correct, computer gets: 0.64 s


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


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 wethrow the switch to position B. How long will it take for thevoltage on the capacitor to be 0.891 V?


Answer:

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


Answer:

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