# EMF in capacitor circuits

In the right circuit in the attached picture, I am given a rectangular loop of wire which is situated so that one end (height $h$) is between the plates of a parallel-plate capacitor, oriented parallel to the field $\vec{E}$. In a text I am using, it is then stated that

"The emf $\mathcal{E} = \oint \vec{E} \cdot dl = 0$ for all electrostatic fields. There is always a fringing field at the edges, and this evidently is just right to kill off the contribution from the left end of the loop. The current is zero."

Question: How is this different to the circuit on the left, where we also a capacitor which has been charged up to a potential $V_0$ at time $t = 0$ and begins to discharge (hence there is an emf and current). Is this not considered an electrostatic field which has a fringing field as well? Why is there an emf and current in the left circuit but not one in the right, what is the difference? Thanks.

The integral $\oint \vec{E} \cdot dl$ is zero for both the circuits.

Case 1

The integration path consists of the wire and the capacitor. Inside the capacitor there is a field, implying the integral is not zero and is the potential difference $V_0$. Hence for the total integral to be 0 , the integral over the wire is $-V_0$. Thus there is a potential difference between the ends of the wire and current flows.

Case 2

This time the entire wire is equivalent to the circuit. For any electric field the integral $\int \vec{E} \cdot dl$ around a closed loop is 0 , and hence for a wire loop in an arbitrary electric field the integral or the potential difference between its ends is 0 implying no current.

The fringing field explanation is given in the text as a lot of people don't consider it and may incorrectly calculate the integral over the wire to be non zero. • Thanks for your answer. Could you elaborate on your reasoning for this statement "hence for a wire loop in an arbitrary electric field the integral or the potential difference between its ends is 0 implying no current." What ends are yo referring to and why would they be zero? – Alex Dec 8 '16 at 11:15
• Electrostatic force is conservative i.e the work done by the electric field depends only on initial and final points irrespective of the path taken. The work done for moving a charge around a closed loop is 0. The integral of E.dl which is the work done per unit charge or the potential difference is 0 for a closed loop. Thus for any closed wire loop subject to any electric field , the integral will be 0. Hence the potential difference between the ends (the same point as the wire is a loop) is 0 . Applying ohm's law V = IR , V = 0 implies I = 0. – cobra121 Dec 8 '16 at 13:53
• Yes but what if we consider in the second circuit (as you did for the first circuit) two points $a$ and $b$ on the wire where $a \neq b$, then $V(b) - V(a)$ is not necessarily equal to zero and by your argument, using Ohm's law we would have a current between $a$ and $b$. What's wrong with this argument? – Alex Dec 8 '16 at 16:52
• Let's say V(a) > V (b) . There are two wire pieces joining a and b (wire is a closed loop). One can say that the current flows in the both the pieces from a to b. By your argument the current will keep flowing without any expenditure of energy which is obviously wrong. – cobra121 Dec 8 '16 at 19:01
• The model I used above is a crude one but extending it further one can say that currents into b cancel each other. Another interpretation can be that positive charge pile up at b and creating deficit at a. These charge then produce a field opposing the original field and bringing the potential difference to 0. If you want to sustain this current one would need to increase the strength of the applied field thus expending energy. This expenditure could be said to be the EMF. Remember EMF itself represents any force which causes charge to flow in a circuit be it chemical or mechanical . – cobra121 Dec 8 '16 at 19:08