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Suppose we have a charged capacitor with charge = $Q_1$ and then an electron beam inside the positive plate of the capacitor.

The electrons entering the positive plate are repelled by the negative plate then the incoming electrons will circulate through the circuit charging the capacitor again at a charge $Q_2 > Q_1$.

So by the relation: $$Q = C * V$$ If $Q$ increases then $C$ or $V$ have to increase proportionally, as $C$ depends on the dielectric and the geometrical characteristics of the capacitor, so I think that the voltage on the capacitors has to increase.
Am I right?

If I'm correct, what happens to the source voltage? Is it maintained or increased or is there a short circuit? enter image description here

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  • $\begingroup$ What happens to the source voltage depends on what kind of power source it is. Is it a power source designed to provide constant voltage regardless of current? Is it a power source designed to provide constant current regardless of voltage? $\endgroup$ – probably_someone Apr 28 at 3:40
  • $\begingroup$ constant voltage regardless of current $\endgroup$ – Ricardo Casimiro Apr 28 at 4:01
  • $\begingroup$ Well, then you know what happens to the source voltage. $\endgroup$ – probably_someone Apr 28 at 4:25
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The red plate in the above diagram is losing electrons (normally), hence it is a positive plate. Likewise, the blue capacitor plate is accepting electrons generated by the battery. Now, P.D (potential difference) is produced due to the difference in the nature of charges(Positive and negative) in the capacitor plates. More the difference in the charges (Eg. +5C and -5C), more is the potential difference available.

Now, in the diagram, you are trying to supply an electron beam to the red capacitor plate, which already contains positive charges, thus in doing so, you are trying to neutralise/cancel out the positive charges on the positive plate, hence decreasing the difference in charges in between two capacitor plates, which ultimately decreases the net potential difference in between the capacitor plates.

Now, normally, when we connect a capacitor in a circuit containing a battery, the capacitor tries to reach the source voltage of the battery, and when it does so, no current starts flowing in the circuit(when its reaches the same value as the source). In your case, the capacitor always tries to reach the source voltage of the battery, and when you supply electron beam to the positive plates of the capacitor, as i mentioned, you are trying to negate the positive charges, decreasing the p.d of the capacitors, now again the capacitor tries to reach the same voltage level of the battery, now, normally, after a long use of battery, the voltage of the battery do decrease, but in your case, at the instant you supply electron beam, it only affects the pd between the capacitor plates and not the voltage source of the battery.

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  • $\begingroup$ So the flow of electrons is produced only by the electron beam or is there also a contribution from the voltage source? $\endgroup$ – Ricardo Casimiro Apr 28 at 4:04
  • $\begingroup$ flow of electrons is mainly produced by the source(battery in your case). That is the main characteristic difference between a capacitor and a battery. Both store electrical energy, but capacitor cannot produce charges unlike battery which it does through chemical reactions. $\endgroup$ – peaceHoper Apr 28 at 4:36
  • $\begingroup$ @Sebastian please mark it as accepted answer if you are satisfied. Click on $\checkmark$ $\endgroup$ – peaceHoper Apr 28 at 12:36
  • $\begingroup$ batteries don't produce electrons... "However, these electrochemical processes change the chemicals in anode and cathode to make them stop supplying electrons. So there is a limited amount of power available in a battery." $\endgroup$ – Ricardo Casimiro May 5 at 2:01

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