So supposedly there is a potential difference between those two rods when the circuit is open. Does this mean theres an electric field present here such that ∫Edl≠0 ?
If I put some random loop of wire in the vicinity (I'm not thinking of Faraday's law, instead just a simple line integral acting on charges), will current start flowing "because theres a potential difference"?
Why does KVL apply to open circuits and how exactly does it?
Is it only true for paths close to the circuit?
Can I take trajectory that goes all the way to infinity and back and still apply KVL?
Can I pick a random trajectory that ends where it started and even if theres a gap, assume that the total voltage drop was the one provided by the voltage source?
Best Answer
KVL can be more generally stated as: the electric field is a conservative vector field. This can be expressed by the path integral: \begin{align} \int_{p1}^{p2} \vec{E} \cdot d\vec{l} = V_{p2} - V_{p1} \end{align} I.E. the integral is "path independent", and the result is the same no matter what path you take (so long as you start at p1 and end at p2). This can be 0, but it doesn't have to be. If \$p1=p2\$ (taking a closed path), \begin{align} \oint \vec{E} \cdot d\vec{l} = 0 \end{align}
This is why KVL even applies: KVL states that in any closed loop, the voltage across it is 0. It doesn't mean that the voltage must be 0 between any two points on that loop. It doesn't matter if there is any conductive path between two points. Having an ideal wire is just an extra statement that two distinct points are at the same voltage.