Log in
Physics
1
answer
0
watching
143
views
23 Nov 2019

As a crude model for lightning, consider the ground to be one plate of a parallel-plate capacitor and the cloud, at an altitude of 550m, to be the other plate. Assume the surface area of the cloud to be the same as the area of a square that is 500m on a side.<BR>a) What is the capacitance of this "capacitor"? (assume air has the same dielectric constant as vacuum)<BR>b) If the electric field strength in the cloud exceeds 3MV/m, there will be a "dielectric breakdown" and charge will flow occur in the form of a lightning flash. What is the potential difference (voltage) between the cloud and the ground at the time of the breakdown?<BR>c) Using the definition of capacitance, determine the charge built up on the cloud at the point of breakdown.<BR>d) If the time it takes for the strike to happen (ie the charge to flow from cloud to ground) is 100

Watch
1
answer
0
watching
143
views

For unlimited access to Homework Help, a Homework+ subscription is required.

Collen Von
Collen VonLv2
23 Feb 2019

Unlock all answers

Get 1 free homework help answer.
Already have an account? Log in
Share

Related textbook solutions

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

Related questions

As a crude model for lightning, consider the ground to be one plateof a parallel-plate capacitor and a cloud at an altitude of 520 mto be the other plate. Assume the surface area of the cloud to bethe same as the area of a square that is 0.30 km on a side.

1) What is the capacitance of this capacitor? Which I found to be1.5 x 10-9 F

2) How much charge can the cloud hold before the dielectricstrength of the air is exceeded and a spark (lightning) results?

1.A capacitor stores charge Q at a potential difference ΔV. Whathappens to capacitance and charge if the voltage applied to thecapacitor by a battery is doubled to 2ΔV?

Is ΔV = Voltage across a capacitor??

2.If I double the space between the plates of a parallel platecapacitor, what happens to the capacitance?

3.If I double the area of the plates of a parallel plate capacitor,what happens to the capacitance of the capacitor?

4. If the voltage across a capacitor is doubled, what happens tothe charge on the capacitor? What happens to the energy stored inthe capacitor?

Is this just the same as #1?



I only grouped these 4 since they are very similar, and #1 and 4may even be identical. I lost my notes and have no clue what todo...

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: ____________

Weekly leaderboard

Home
Homework Help 3,900,000
Physics 110,000
Start filling in the gaps now
Log in