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In this image one plate of C1 is at potential of 100v and one plate of C3 is earthened. Which type of circuit is this? If it's open, why would the charge flow?

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  • $\begingroup$ For a circuit to be closed, you need a difference of electrical potential, not actual closure of the lines. $\endgroup$ – Grego_gc Feb 18 '20 at 15:17
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A circuit does not need to be closed for current to be able to flow though it. Current arises whenever there is a potential difference applied on the ends of a conductor. Here, the point at $100V$ is at higher potential as compared to ground which is at a lower potential of $0V$. You might as well join the two open ends with a battery of constant emf of $100V$.

However note that as there are capacitors connected, the current flowing through the circuit will eventually become zero.

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The circuit is open in the sense that there is no continuous conductive path for charge to flow, since electrons can't physically cross the space in between the plates of the capacitors. However, this does not mean that charge can't flow in the conductive paths that are present in the circuit causing the movement of charge on to and off of the capacitor plates. But the duration of the flow of current will be limited to the time where voltages across the capacitors are changing since the relationship between current and voltage for a capacitor is

$$i_{C}(t)=C\frac{dv_{C}(t)}{dt}$$

So current will only flow in the circuit as long as the voltages are changing across the capacitor plates, and those voltages are only changing while the capacitors are charging. In the case of a 100 volt dc source (e.g. a battery), once the sum of the voltages across the series capacitors equals 100 V potential source, current ceases.

Your diagram is missing some critical information to complete the analysis. Is the 100 volt source a constant (dc) voltage source, like a battery mentioned in @Sam answer, or an ac source of what frequency? How long have the circuit conditions shown existed? If the source is dc and the circuit has existed a long time, the current will be zero since voltages will all be constant. On the other hand, it the circuit shown is the instant following closing of a switch to the voltage source, charging current will exist limited only by the resistance that is necessarily in any real circuit.

Hope this helps.

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Charge will flow through the conductors when the pd is first applied, but after a short time (dependent on the inevitable small amount of resistance in the circuit) it will, to all intents and purposes, stop flowing. Equal and opposite charges will then reside on the plates of each capacitor.

Before the pd is applied, the circuit is certainly open. After the pd has been applied (presumably by connection of battery or power supply), I think an engineer would say that the circuit is closed, but I'm uneasy about this usage, since none of the capacitors can conduct. In short, I think that it is not useful to describe the circuit as open or as closed.

Instead we can simply say that charge flows because of the application of the pd, which produces electric fields in the wires. As the capacitors charge the pds across them increase and those across each wire diminish. Eventually there are no pds across the individual wires, and the applied pd is equal to the sum of the pds across the capacitors.

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