🏷️ LIMITED TIME OFFER: GET 20% OFF GRADE+ YEARLY SUBSCRIPTION →
Log in
HomeStudy Guides420,000US230,000

PHYS 101 Midterm: UIUC PHYS 101 101 FA08 HE1ABlank

17 views4 pages
turquoisegnu128
7 Mar 2019
School
University of Illinois
Department
Physics
Course
PHYS 101
Professor
All

For unlimited access to Study Guides, a Grade+ subscription is required.

Document Summary

P. avery, a. rinzler, s. hersh eld: in the figure the capacitances are in micro-farads ( f). The equivalent capac- itance of the network of capacitors is: Solution: the 6 f and 12 f capacitors in parallel have capacitance 18 f . These capacitors are in series with the other 6 f capacitor to the e ective capacitance is 4. 5 f : in the previous problem, the charge on the positive plate of the top 6 f capacitor is: Solution: 12 volts across an e ective 4. 5 f capacitor gives a charge of cv = 54 c. The 6 f and 12 f capacitors in parallel have e ective capacitance of 18 f so the voltage across them is q/c = 54 c/18 f = 3v . When charged to 6. 0 v the potential energy stored in the device is: Solution: the capacitance of the parallel plate capacitor is c = oa/d.

Get access

Grade+20% off
$8 USD/m$10 USD/m
Billed $96 USD annually
Grade+
Homework Help
Study Guides
Textbook Solutions
Class Notes
Textbook Notes
Booster Class
40 Verified Answers

Related textbook solutions

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

Related Documents

PHYS 102 Lecture 11: Capacitor in series
amaranthtiger876
ECT 2440 Chapter Notes - Chapter 3: Capacitor Types, Dielectric, Oliver Heaviside
limehyena975
ECE-105 Chapter Notes - Chapter 3: Linear Variable Differential Transformer, Capacitor Types, Polypropylene
limehyena975

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:

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: