- What initially causes the capacitor to accumulate a charge difference? I would expect the current to see this as an open circuit, given that no current flows through the dielectric to ground, and therefore there would be no initial flow.
- When we discuss a build-up of positive charge on one of the plates, we are saying the valence electrons have been repelled by the negative charge of the the other plate. Maybe this is a chicken/egg situation, but shouldn't the electrons on the negative side of the plate be repelled by the same force (even more since it's closer) as the positive side, inhibiting the charge buildup from occuring in the first place?
- In the picture below, won't the discharge reverse the direction of current? That is, it's going to flow from the negatively charged plate, across the LED, resistor, gate, and then to the positively charged plate?
- When the capacitor discharges, the current seems to want to flow to redistribute/equalize charge across the circuit. This is different than just flowing to ground (or the positive terminal on the battery), as I am used to seeing. If when the gate closed to allow discharge there was also a path to the positive terminal on the battery, would this change anything?
closed as too broad by Eugene Sh., Voltage Spike, uint128_t, Dmitry Grigoryev, jonk Dec 23 '16 at 10:15
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Current does flow through the capacitor, through the dielectric. But only while it's charging.
This discussion of capacitors may help. The charge buildup does repel further charge buildup. That's why the rate of current flow decreases as the capacitor charges. Work must be done on the charge to move it - the integral of that work is the energy stored in the capacitor.
Yes. The animation is very bad at showing this.
Yes, current will tend to flow so as to equalise the potential on both sides. Neither is necessarily related to ground. That's how a Cockroft-Walton multiplier works.
shouldn't the electrons on the negative side of the plate be repelled by the same force (even more since it's closer) as the positive side...
Electrons flow from the battery to build up on the negative plate. This forces electrons away from the positive plate where they flow out the lead and to the battery (+), which is collecting them.
...repelled by the same force as the positive side, inhibiting the charge buildup from occurring
This is what happens when the capacitor is charged. Since the dielectric is an insulator, no electrons flow through it; they just build up until the force from the battery is balanced by the force of the crowd of electrons pushing back.
This is different than just flowing to ground
"Flowing to ground" is a common lazy way to overlook the fact that the current ultimately must flow back to its power source--in this case, the battery, or the capacitor plate. Connect a ground symbol to the bottom of the circuit, and now we can talk that way. But when we get this specific about where the currents actually go, it's more accurate to identify where they go, not just "to ground".
https://goo.gl/8p1BZf < JAVA Falstad Simulator
Consider charging a Cap like jump starting a dead battery. ( a tiny one) A battery is like a Cap but thousands of times more charge storage in Farads.
Do you understand that rate of change of charge flow in milli-Coulombs/second = milliamps. Since Battery has very low series resistance e.g. 10 milliohm for LiPo and some E-caps are similar but much lower capacitance per unit volume compared to a battery.
I assume you understand polarity of current is not the flow of electrons for electronics by convention rather from positive to negative for logical reasons not physics.
Scope traces show max:min only for each node. So Cap shows a steady -ve drain and large peak pulse. such that if ideal Ichg * t integral area under the charge spike = steady drain to LEDs Id * t. for conservation of charge. But I added realistic Rs series values for each part. Since Red has a lower forward voltage it illuminates longer.