Question

Hello there! I have a question. Kirchoff's first Law states that the line integral of E.dl on a closed loop is zero. I also have seen that this line integral is zero only when we are dealing with conservative electric fields. So why do we consider that is zero when we resolve circuits with batteries and resistors, isn't the electric field produced by a battery non conservative?

Answer 1

An electric field is conservative if the value of the line integral of E.dl is independent of the path you choose to follow, and this is true with a simple Battery/Resistor circuit.

Things get more complicated when there is changing magnetic flux through a circuit, because additional EMFs are added to the circuit, but they are very dependent of the path. Kirchoff's first law only holds when there is no changing magnetic flux through the circuit.

If you haven't studied this yet, all of this will make sense when you see Faraday's law.

Answer 2

The Kirchoff law states that because we're talking about electrostatics fields so the field must be conservative thats why :
\[\sum_{i=1}^{n} AV=0\]

Answer 3

I got confused because I was taught in highscool that the electric field produced by a battery was non-conservative. Now I know It wasn't right at all!

Answer 4

Kirchoff's laws are not talking about electrostatics - how would there be any potential difference in a static case? There are currents in these circuits precisely because there are differences in potential from point to point which are associated with electric fields maintained by batteries/changing magnetic fields in the conductors - that's definitely not static

Answer 5

How much work is required to increase the separation of the plates from 2.29 mm to 6.87 mm?

Answer 6

How much work is required to increase the separation of the plates from 2.29 mm to 6.87 mm?

Answer 7

i am having trouble getting that question. any help