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I have some confusion about the negative terminal of a capacitor. Consider the following

enter image description here

Now, my understanding was that postive charges accumulate at the positive ("top") plate of the capacitor, setting up an electric field within the capacitor which causes negative charges to accumulate at the bottom plate.

Does that mean that the bottom plate of the capacitor will be at -5V eventually (at steady-state). Or does it mean it will still be 0V? The circuit simulation tool I ran this with, says it is 5V.

Can someone explain what the negative plate voltage is? And also what happens if there was no ground but another circuit component there?

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Let's go back and try to understand what voltage is. Let's say you have a 1.5 volt battery. What is the voltage difference between the positive and negative terminals - 1.5V of course. Now what is the voltage of the negative terminal with the battery floating in space, or the positive terminal? We don't know. Voltage is a potential difference between 2 points. Ground is a reference point. You could tie either battery terminal to ground and it is still a 1.5V battery. In your circuit you could tie the positive side of the capacitor to ground and leave the negative side open. You still have 5V across the capacitor but the positive side would read 0V and the negative side -5V. So remember that a "ground" point is a measurement reference. You could tie the reference to earth or leave it open but that's another topic.

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The negative plate of the capacitor is connected to ground. Therefore, if you ask for the voltage at that single point (rather than explicitly with respect to some other point) then the answer must be 0V. This point is always at 0V, by definition, because it is connected to ground.

You are correct that the electric field on the capacitor causes charge to flow from the negative plate to ground. The amount of charge exiting from the negative plate is exactly equal to the amount of charge that enters the positive plate, so the entire capacitor structure remains charge neutral. As voltage increases across the capacitor the voltage across the resistor decreases, which means that the current must also decrease. Given enough time, the voltage on the capacitor rises to be the same as the supply voltage. At that point the voltage across the resistor falls to zero and current flow also falls to zero.

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  • \$\begingroup\$ Great explanation. This leads to me follow-up question - what if the bottom plate of the capacitor isn't connected to ground? For example in a 555 timer circuit. \$\endgroup\$ – AlfroJang80 Sep 5 at 0:24
  • \$\begingroup\$ No one can give you an answer unless you tell us where the bottom plate is connected. \$\endgroup\$ – Elliot Alderson Sep 5 at 11:37
  • \$\begingroup\$ electronics.stackexchange.com/questions/456224/… \$\endgroup\$ – AlfroJang80 Sep 6 at 17:38
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If the bottom plate is not grounded, then you must clarify what is the "reference" point.

For example, the (+) plate may have, say, +3V with respect to the (-) plate.

That would be the same as saying the (-) plate has -3V with respect to the (+) one.

So, what would be the voltage at the (-) plate with respect to ground? To answer that, it would be necessary to know the circuit connected to that plate.

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