# Meaning of potential in a discharging capacitor

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.

• $\mathbf{E}$ = $-\nabla \phi$ - $\partial_{t} \mathbf{A}$, 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. – honeste_vivere Jan 5 '15 at 19:16