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I've tried to research this answer for quite some time but i'm struggling to find a proper answer. I think my question falls between antenna theory and batteries.

So in a dipole antenna we supply a sine wave across the antenna and this generates a small current due to capacitance and the electron flow at that frequency.

Now suppose we take this example to a DC circuit I understand DC current cannot flow in an open circuit because the battery has a net neutral charge and if you just discharge one side the other side creates an electric field to pull the discharged side back.

But if put that battery across two huge metal blocks one on each terminal will you get a current flow as one side has electrons added and one side electrons removed? As I understand for one side to be at +12v it must have a lack of electrons and the other side a surplus so there must be some electron flow to make each block equal to the voltage? Otherwise my understanding of the antenna doesn't seem to make sense.

Then theoretically if you charge one block to 12V and one block is at 0v then disconnect the battery what happens to the charges on each block?

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    \$\begingroup\$ FYI, sometimes we talk about the transient behavior of a circuit. That is, what happens when we suddenly change something (like, what happens when we switch on the power.) Other times, we want to talk about the steady state behavior of a circuit. The rule that says, "current cannot flow in an open circuit" belongs to the realm of steady state analysis. A capacitor is an "open circuit," and a capacitor can't conduct steady DC current, but you'll get a big spike of inrush current (a transient) when you first connect your capacitor to a DC power supply. \$\endgroup\$
    – Solomon Slow
    Commented Aug 6, 2019 at 18:54

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There will be a brief flow of current when you first connect the battery to the metal blocks. That current will charge the capacitance between the blocks. Once that capacitance (and other stray capacitance) is charged, there will be no further flow.

If there is no leakage current between the blocks, the charge on the capacitor formed by the blocks will remain.

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  • \$\begingroup\$ Do you regard the self-capacitance of each block as "stray"? \$\endgroup\$
    – glen_geek
    Commented Aug 6, 2019 at 16:08
  • \$\begingroup\$ @glen_geek, Stray capacitance is capacitance. Whether or not you choose to say "stray" depends mostly on whether it causes a problem that the circuit designer failed to anticipate or not. From the physics viewpoint though, it's just capacitance. \$\endgroup\$
    – Solomon Slow
    Commented Aug 6, 2019 at 18:58
  • \$\begingroup\$ Odd that this answer never got any rep. I suppose nobody ever saw it, even OP. \$\endgroup\$
    – Simon Fitch
    Commented 2 days ago
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Your two metal blocks (assuming they're insulated from each other) make up a capacitor. Yes, current will flow as the capacitor charges. Then when you remove the source, the charges will find some path, however high the resistance, and the capacitor will discharge at a rate determined by the capacitance and the resistance.

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  • \$\begingroup\$ Odd that this answer never got any rep. I suppose nobody ever saw it, even OP. \$\endgroup\$
    – Simon Fitch
    Commented 2 days ago

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