If both the plates of a capacitor are connected to positive voltage (say, one is +10V and the other is +5V), but there is a voltage difference between the plates, will the capacitor be charged? And why?

  • $\begingroup$ See capacitance coefficients aka coefficients of potential: en.wikipedia.org/wiki/Coefficients_of_potential. Your conceptual difficulty stems from the implicit assumption that there are only two plates, when in reality there are at least two more: the connectors of the power supply. $\endgroup$ – CuriousOne Dec 31 '14 at 10:19
  • $\begingroup$ Unclear: if they're connected to the same voltage, how could they possibly be at different potentials? $\endgroup$ – Carl Witthoft Dec 31 '14 at 13:46

About two points with positive voltage

Remember that voltage is simply difference in electric potential (which is electric potential energy per unit charge).

Like for gravitational potential energy, you can define your reference point anywhere. You can define zero potential to be one of the points, you are measuring between. Or to be the ground. Or another point in the circuit. It doesn't matter, since all you are looking for is the potential difference.

View electric potential as pressure as an analogy. If there is one bar on one side and two bars on the other side, there is a substantial difference and air will start moving. If there is 100 bar on one side and 101 bar on the other side, this still gives the same result. Only the difference is important.

In your case, the two positive voltages you find must be because they are measured with a third reference point. You can have any potential at those points by choosing another third reference.

Bottom line is: if they are both positive doesn't matter. Only the difference matters.

Will the capacitor be charged in your case?

That surely depends on the circuit. Yes, with a potential difference (a voltage) between those two plates (not regarding their individual potential in reference to some other point) charges will flow and the capacitor will be charged - unless of course it is fully charged already, or the circuit is not finished.

Capacitance of a capacitor is found with:


So for a set voltage, Q is the amount of charge that will build up. When this is reached (or nearly reached) the capacitor is fully charged.


If you connect each plate to a DC supply, you will either get zero charge on the plate if the DC voltages are the same in both magnitude and polarity, or you will get a negative charge on one plate and a positive charge on the other if the DC voltages differ at all.

So, say we have two DC voltages, $V_1$ and $V_2$, where $V_1 > V_2$, that we connect each terminal to. Even if both DC voltages are positive, or both are negative, this will set up an electric field in the circuit that drives the current from the higher voltage to the lower voltage, i.e. Move electrons from the lower potential to higher potential.

So, on the $V_1$ side of the capacitor, electrons will flow from the plate to the voltage terminal, and on the $V_2$, electrons flow from the terminal to the plate, so the two plates become charged positively and negatively respectively. Electrons flow until the voltages of each plate equal that of the connected voltage.

So, when current stops flowing, there is a now a potential difference across the plates as well as a charges of equal magnitude and opposite polarity. This potential difference is equal to $V_1 - V_2$. The magnitude of the charge on each plate is related to this potential difference by the following equation:

$$Q_{plate} = C(V_1 -V_2)$$

This tells us that if the two terminals have the same voltage, the capacitor will have no charge after the transients die out over time (when electrons are moving). Otherwise we get a charge proportional to the difference in the voltages, of opposite polarities on the two plates.

Just to highlight that the fact the two voltages are positive doesn't matter: voltages (i.e. Potentials) are relative, so if you increase both voltages here, you get no change in the circuit's response. Try putting values of voltage into the equation above, and repeat for voltages that are offset by the same amount: the charge does not change!

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  • $\begingroup$ ** Will the electrons move back to V_2 because V_2 is positive and the plate is negatively charged? **<br/></br>The answer is no. The other plate is positively charged. So it will attract the electrons of the negatively charged plate. Moreover, as the positively charged plate has positive electric potential (for its positive charges), it will have influence over the other plate too (because they are too close) and vice-versa; And the net electric potential on both the plates will eventually be zero. $\endgroup$ – arpon Dec 31 '14 at 14:08
  • $\begingroup$ No, the electrons will slow down until they stop. It is important to recognise that the polarity of a voltage potential has no physical significance like that of the polarity of charge. If a voltage is positive, that only tells you that it is higher than an arbitrarily selected reference voltage. Steeven's answer covers that rather nicely. It is misleading to think of electrons going from negative to positive, but that electrons flow from lower (not necessarily positive) voltage potentials to higher potentials (opposite direction of current). $\endgroup$ – Involutius Dec 31 '14 at 14:17
  • $\begingroup$ And, in the end, each plate gains the same voltage as the DC connection, so if the voltages are equal along a wire, no net electron flow is present. $\endgroup$ – Involutius Dec 31 '14 at 14:26

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