Parallel Plate Capacitor Physics Problems

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This physics video tutorial provides a basic introduction into the parallel plate capacitor. It explains how to calculate the electric charge stored on a capacitor given the capacitance in farads and the voltage across the capacitor. It explains the meaning of the farad in terms of coulombs per volt. In addition, it discusses how to calculate the capacitance, electric charge, and the electric field inside a parallel plate capacitor given the separation distance of the plates and the area. Finally, this video discusses how a capacitor works. it explains how it can store a charge and then release it for later use.

Physics 2 - Basic Introduction:

Coulomb's Law and Electric Force:

Electric Fields:

Electric Potential:

Electric Potential Energy:

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Parallel Plate Capacitor:

Energy Stored In a Capacitor:

Energy Density of a Capacitor:

Dielectrics of Capacitors:

Capacitors - Review:

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Electric Current In Circuits:

The Electric Battery:

Ohm's Law Problems:

Resistance and Resistivity:

Physics PDF Worksheets:
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cleopatre
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I'm an industrial engineering student in Turkey and i like your all videos. They are really useful and you are doing really good work. Thank you so much for everything 🌸

berilbulut
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Hello. I believe your value for Capacitance is incorrect (7:30). You show it's (2.655 * 10^-9), but if you use the numbers used in the problem you actually get (3.0 * 10^-9).

Really apprecaite all these videos. I'm anxiously waiting to graduate with EE, and hopefully be able to support you financially. You're responsible for creating so many future engineers and professionals. I truly hope that we come back and support you when we're no longer broke college students.

MYGGWarcraft
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Thank you for the easy and simple explanation. Helped me clear some doubts ✌️

nayeem
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mertdunver
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These videos explain it better than my textbook

mylitomutalisk
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When I hit the lottery one day I will track you down and hand you a very large sum of money. Thank YOU!!

jstroner
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A parallel-plate capacitor of 20 𝜇𝐹 capacitance is charged to a potential difference of 5 V. How much charge is
stored in each plate?

jakecanda
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For d I used 5mm --> m which is 0.005, however my answer came out to 10^-10 as opposed to -9. So what step am I forgetting here, I know the conversion is correct, but I don't understand why .5*10^-3 was used in its place.

lilcody
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so is that area in the equation just for one plates, or both of the plates?

IamKudos
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Thanks for your knowledge. What software do you use for writing?

ronaldos
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Thank you very much, I owe you a coffee for a real :)

jahansaid
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before this vid, i thought F was ferrets, instead of farads

hi.
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can i know why did you use k as 1 instead of 9*9(pow9)?

fadlyahmad
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wait in the end (9v)/(0.005m) is 1800, not 18000
did i do something wrong here?

ukoqtob
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But they asked for charge on each plate?

dharanishree
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