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A typical case

Staring with facts, 1)For a current to flow there must be a closed loop.

2)For a capacitor to be charged there must be a potential difference across it,

3)And obviously a charge must be transferred from some source to the capacitor while charging.

Considering a typical case as in the above picture,

1)How can a current flow from the battery to any one of capacitors to get it charged without a closed loop??

(It leads to an heavy confusion that the space between the capacitor doesn't conduct any current, Then how does the battery get triggered to release the charges??)

2)In capacitors C¹ and C² the charges are supposed to be equal as many textbooks regards, but why and how could it be??

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  • $\begingroup$ your first statement :Staring with facts, 1)For a current to flow there must be a closed loop. is wrong. you can carry a large from one charged body to an uncharged. only if a current has to flow for a long time you have tp have a closed loop. did you hear about influence? $\endgroup$ – trula Sep 25 '19 at 10:45
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1)How can a current flow from the battery to any one of capacitors to get it charged without a closed loop??

First of all, if the circuit shown in your diagram has existed for a long time there will be no current flowing. That is because the voltages across the capacitors will not be changing in time. The current in a capacitor is given by

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

If the voltage across the capacitor is not changing in time, there will be no current. The capacitor looks like an open circuit in dc circuits after any transients (time changing voltages and currents) have died out, which is the situation after a long time.

But if the capacitors originally start out not charged, or not fully charged to the potential of a battery, current will flow once switched onto a voltage source. The battery pulls charge from one plate and pushes that same charge onto the other plate, creating a net positive charge on one and negative charge on the other. To do this the battery does work. The energy of that work is stored as electrical potential energy in the electric field between the plates as well as dissipated as heat in resistance that is always present in a circuit.

It is this process of pushing and pulling charges that results in current flowing in the conductors connecting the capacitors to the batteries, even though though charge does not actually move between the plates within the capacitor.

2)In capacitors C¹ and C² the charges are supposed to be equal as many textbooks regards, but why and how could it be??

Since the capacitors are in series, when the capacitors are switched onto the batteries the same current (charge per unit time) flows to each capacitor. Since the same charge flows per unit time to each capacitor, the plates of the series capacitors acquire the same amount of charge, regardless of the value of capacitance. And that charge will be

$$Q_{1}=Q_{2}$$

$$C_{1}V_{1}=C_{2}V_{2}$$

From KVL:

$$E_{1}-V_{2}-E_{2}-V_{1}=0$$

So the voltages and charge can be calculated.

Hope this helps.

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  • $\begingroup$ Your statement, "The battery pulls charge from one plate and pushes that same charge onto the other plate, creating a net positive charge on one and negative charge on the other" was so satisfying! But one thing I noticed is, your explanation directly relates why Sparks fly over two wires connected across a high voltage battery! That was truly amazing. $\endgroup$ – Mukunth A.G Sep 25 '19 at 17:05
  • $\begingroup$ But, in that typical circuit that I had mentioned, can you improve my intuition on pulling of charges on a single capacitor by two different batteries in two directions? $\endgroup$ – Mukunth A.G Sep 25 '19 at 17:14
  • $\begingroup$ @MukunthA.G I'll give it a try. Let's put an open switch in the circuit. Let's also assume both capacitors initially have no net charge on the plates. You then close the switch. You know that if the two battery voltages are equal in magnitude there will be no current flow since they oppose each other with equal and opposite voltages. So we need a voltage difference to charge the capacitors. $\endgroup$ – Bob D Sep 25 '19 at 19:53
  • $\begingroup$ Now suppose E2>E1. We will have a clockwise current of positive charge. Now in reality you need some resistance in this circuit, otherwise the initial charging current will be very large. Capacitors have been known to burst with a loud pop without adequate circuit resistance to limit the maximum charging current. By superposition, we can consider the effects of each battery separately imagining the other shorted, then add algebraically add the effects. $\endgroup$ – Bob D Sep 25 '19 at 19:53
  • $\begingroup$ Consider battery E1. It pushes positive charge onto the bottom plate of C2. That positive charge pushes an equal amount of positive charge off the upper plate due to the repulsive force between like charges, giving the top plate an equal and opposite net negative charge. At the same time, the negative terminal of E2 is pulling positive charge off of the bottom plate of C1, making it negative. That attracts positive charge that left the upper plate of C2 to the upper plate of C1, making it positive $\endgroup$ – Bob D Sep 25 '19 at 19:54

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