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Recently I have learned about the behaviours of capacitors are as follows:

  1. When connected to a voltage source the voltage on the the plate on the positive side begins to rise slowly until it reaches the same voltage as the supply, while the other plate, which is connected to ground, remains at 0v.

  2. Capacitors have the tendency to keep the voltage across two plates the same, and therefore if one is 10v and the other is 0v, and the 10v is suddenly pulled to ground, the other plate will be pulled to -10v to maintain the voltage differences.

These two statements seem contradicting to me. Specifically, if capacitors tends to keep plates voltage the same, then why when it charges, the other plate remains 0v? And unlike voltage rising(charging) which happens gradually over time, why does the plate voltage change from 10v and 0v to 0v and -10v instantaneously when pulled directly to ground, that is, why in this case it doesn’t lower its voltage gradually over time?

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    \$\begingroup\$ I feel like this is another case of people confusing voltage for an absolute quantity. It isn't; it's a relative quantity. \$\endgroup\$ – Hearth Oct 15 '19 at 13:08
  • \$\begingroup\$ I don't hold either of your statements to be true. I recommend going back to a physics or circuits text for better descriptions. \$\endgroup\$ – Scott Seidman Oct 15 '19 at 13:14
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    \$\begingroup\$ The plate connected to ground cannot become -10 V, if it would then ground would become -10V as well since they're connected. But ground is by definition 0 V. The -10V will only happen when there is no direct connection, for example a connection through a resistor. \$\endgroup\$ – Bimpelrekkie Oct 15 '19 at 13:15
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    \$\begingroup\$ Learn this instead \$i_c = C\cdot \frac{dv}{dt}\$. That tells you everything you need to know about an ideal capacitor. \$\endgroup\$ – Andy aka Oct 15 '19 at 13:16
  • \$\begingroup\$ If one plate is directly grounded then that plate will have a very hard time trying to change in voltage in response to the other plate's voltage change. However if there is reasonable resistance in the circuit then you will start to see some of the effects that you listed. You have a lot more studying to do, good luck. \$\endgroup\$ – Nedd Oct 15 '19 at 13:20
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When connected to a voltage source the voltage on the the plate on the positive side begins to rise slowly until it reaches the same voltage as the supply, while the other plate, which is connected to ground, remains at 0v.

This is basically correct, but remember that all voltage is relative and the capacitor doesn't "know" where ground is. It acquires an electric field with a potential difference across the plates.

Capacitors have the tendency to keep the voltage across two plates the same, and therefore if one is 10v and the other is 0v, and the 10v is suddenly pulled to ground, the other plate will be pulled to -10v to maintain the voltage differences.

Careful. I'm guessing that this experiment isn't connecting 0V to both plates at the same time, but instead "floating" the capacitor by disconnecting both plates. Since it's disconnected, there is no longer a circuit, and its voltage relative to ground can't be sensibly measured.

So you measure +10V from the "top" terminal to the "bottom" terminal while it's floating. Connect a wire to the "top" terminal, and you still measure +10V top to bottom. Because voltage is relative. If you label the top wire "0V", then yes the bottom terminal appears at -10V.

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    \$\begingroup\$ Point 1 is incorrect - when it connects to a voltage source, a massive current flows to raise that plate immediately to the voltage of the voltage source. \$\endgroup\$ – Andy aka Oct 15 '19 at 14:12
  • \$\begingroup\$ @Andyaka Only if it is an ideal voltage source with zero internal resistance and a theoretically perfect capacitor. The current is proportional to the total loop impedance, which includes the internal resistance of the voltage source, any wiring resistance and any series inductance. The internal construction of the capacitor will also have an effect. \$\endgroup\$ – Peter Jennings Oct 15 '19 at 14:20
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    \$\begingroup\$ @PeterJennings so, you should decide... does the voltage on the plate rise slowly as the OP (and pjc50) "believe" or does it rise immediately when connected to a voltage source. What side of the fence would you choose to sit on? \$\endgroup\$ – Andy aka Oct 15 '19 at 14:43
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    \$\begingroup\$ @Andyaka As I said, it depends on whether you are talking about perfect textbook voltage sources and capacitors (= instantaneous change) or real world components that you could actually perform this experiment on. Modern, good quality components and voltage sources can be close to perfect, but you can never achieve infinite currents and instantaneous voltage changes. \$\endgroup\$ – Peter Jennings Oct 15 '19 at 17:37
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When connected to a ideal voltage source, an ideal capacitor immediately goes to the voltage of that source. There are no such things as ideal voltage sources, and there are no such things as ideal capacitors. There are always effective series resistances. Also, one plate is at ground by definition, if it is attached to ground.

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There is also no idea connection to ground. If there was an ideal connection to ground, and you assumed Earth was ground, and Earth was also ideally conductive, then the entire Earth would become the "bottom" plate, and the electrons pumped into the Earth when the top plate was positively charged by some current source connected between the top plate and the planet would lower Earth's total charge by some near infinitesimal unmeasurable amount.

In actuality, the actual charge and thus voltage on the Earth's surface varies by many kiloVolts over time and location due to being not ideally conductive, plus ongoing atmospheric, radiative, and cosmic effects. (Lightning!). So, your -10 Volts is lost in the "noise".

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