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I am dealing with this thing I cannot figure out. When a capacitor is discharging, the electric field inside it varies with time so we cannot perform the line integral to determine the potential difference between the plates. Then, potential differences has actually no meaning. So what does a voltmeter measures in this case? The only easy solution I can think of is that there's the assumption that electric field changes so slowly compared to the velocity of moving charges that potential difference has still meaning.

Thanks in advance.

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  • $\begingroup$ $\mathbf{E}$ = $-\nabla \phi$ - $\partial_{t} \mathbf{A}$, in general, so I am not sure why you argue that a time-dependent electric potential has no meaning. $\phi$ can equal $\phi(t)$, there is nothing wrong with that. Besides, isn't this just an RC or LRC circuit, depending on how you set it up? Then the result is just an exponential decay/growth versus time. $\endgroup$ – honeste_vivere Jan 5 '15 at 19:16
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In the time dependent problem there is also a magnetic field due to the displacement current which will cause a force on a moving charge/current. In that sense the potential is not defined. What a voltmeter will measure will depend on how the wires are running around the capacitor. Imagine you make a couple of turns around the capacitor body. The time varying magnetic field will now induce an extra voltage in this coil. That's the thing with non-conservative forces: they depend on the path and not just on two points.

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