Changing the Capacitance of a Capacitor

[@roderickjamesbeaumontfowle8824]

You always manage to show a far more intuitive way of analysing and solving problems, even when I thought I had it all down pat. Thank you.

[@jelebalxx8782]

you saved my midterm thanks for these videos you're the best!!

[@galayepaez237]

What is the effect on a charge if we increase the distance between the plates of a parallel plate capacitor?

[@lasseviren1]

If the capacitor is connected to the battery when this is done the charge on the capacitor will go down. Since C = k(epsilon naught)A/d , increasing d decreases C. You must do work to pull the plates apart (since they are oppositely charged) and that work transfers energy from the capacitor to perhaps the battery. If the capacitor is instead disconnected from the battery, the charge will likely remain the same since it has nowhere to go.

[@TedSamore]

8:50 Does that mean that there is a 4 voltage drop over the disconnected battery? Thanks !!

[@bgediktas9435]

Your videos are awesome!

[@anti-kafir-t8r]

AS BAYRAKLARI AS AS AS!!!111!!!

[@seravenerdi]

How do I reduce a 2400pF capacitor into 1600pF ?

[@lasseviren1]

I'm not sure if you're asking to solve a physics problem or if you really want to change the capacitance for a practical reason. If it's for a physics problem, then increase the distance between the plates or change the dielectric. C=k(epsilon naught)A/d. Of course that's not a practical option if you really want to do this. Then you might put a second capacitor in series with the first, 1/Ceq = 1/C1+1/C2. Still not easy to do in a more complex circuit. Not trying to be a wise guy, but if you really want to do this for a real circuit, buy a 1600pF capacitor.

[@cdelcarmen90]

Thank you so much for your clear explanation.

[@elliottmarcel1506]

sick video!

[@bluegrapes2010]

this is so so helpful!!! Thanks!

[@onzku]

This was really helpful! Thanks man

[@waheedx97]

god bless your soul!

[@BatusaiJack]

GOLD