Why does a charged capacitor provide current to a circuit?
View Single Post
Jan2-13, 01:41 PM
The external field is negligible only as compared to internal field. And that has to do with the fact that path along external field line is much greater than path along internal field lines.
OK, I get it. This is because, for a capacitor with a relatively large capacitance, the plate separation is taken to be small. Right?