If I have a single metal plate, a battery, and one wire. If I connect the wire to the positive on the battery to the metal plate, does that remove electrons from the metal plate?

I am attempting to understand capacitors better. I know what they do, and mostly what's going on at a lower level, but I lack fundamental knowledge. Do I understand this correct about the single, metal plate? No circuit here; no capacitor here (I think). Just a battery and positive via a wire touching a metal plate.

I could not draw just a metal plate, so I used a capacitor symbol.

I thinks this really answers it, How does a capacitor block DC?

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Same with the other way,

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  • \$\begingroup\$ Can you show what you mean with a schematic? \$\endgroup\$ – winny Oct 4 '18 at 21:24
  • \$\begingroup\$ @winny I added one. \$\endgroup\$ – johnny Oct 4 '18 at 21:37
  • 3
    \$\begingroup\$ Assuming the ideal case where there are no unknown connections, the answer is no outside of a moment of equilibration as the charges redistribute (if they need to.) But this is no different than two pieces of metal and no battery, at all. It only takes just a few charges, by the way, to cause quadrillions of electrons per second to take the bend in a wire. So charges are very powerful. Most of us don't fully apprehend just how powerful. \$\endgroup\$ – jonk Oct 4 '18 at 23:32

Maybe a handful, but definitely not enough to do anything besides equalize the potential (voltage) between the two.

Basically, the only thing that happens is charge balance occurs and the voltage becomes the same between the battery's terminal and the plate.

An extreme example of this is when helicopters hook themselves to the high voltage power lines. There's actually a considerable amount of electrons flowing, but it's not harmful and it's only temporary until the potential is balanced. In this case, the battery is the power line and the plate is the helicopter.


For an ideal circuit? Such as analyzing schematics? The circuit is open. The current remains zero. No charges move. Resistors, capacitors, inductors, these are all two-lead components, and if they are not part of a network loop, no charges can move at all. That's the basic theory we need to learn as beginners.

Now in the real world, the air is not an insulator, but acts as capacitor-dielectric, and also as 10^30 ohm resistors because of cosmic ray ionization, Radon, etc. The capacitor will charge slowly through the "open" circuit, unless the capacitor's dielectric has far lower resistance than the leakage resistor through the open circuit.

Also, the entire circuit is covered with parasitic capacitors. In your schematic, draw some 0.1pF capacitors between all circuit nodes. Also add more of them between every circuit node and the earth. Now, when your capacitor is sitting on the table, it has tiny capacitors connecting both leads to earth-ground. The battery does too. There are also tiny capacitances between the battery leads and the capacitor leads. If you now bring them near, and touch only one terminal together, you're suddenly creating capacitive voltage-dividers.

Then, if your capacitor value was well above 100pF, the tiny parasitic capacitors will be insignificant. No charges move. (If any actually do, well, they're part of voltage-shifts which are way below 1%, and ignored. Also, your entire capacitor becomes slightly charged wrt earth-ground: both plates have excess charge, but no volts between plates.) On the other hand, if your capacitor was 0.5pF, then parasitic capacitances between every wire and earth-ground will seem quite large. Touching one side of the capacitor to one side of the battery will charge up the whole capacitor.

Also see my 1998 bit about Engineer's Capacitor, the solid metal sphere with the microscopic cleft.


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