# Strange behavior when connecting ideal voltage-source and capacitor in series?

I just learn from youtube video that if capacitor is connected with ideal voltage source in series, its voltage graph would be like this :

(Sorry, but adding as picture is much better than drawing it by myself)

What I can't understand is that when Vin goes high, why does Vout goes high, too? One side of capacitor could go high because voltage source draw electrons from capacitor, but other side of capacitor is floating state!..

I have no idea how I can interpret it.

Thanks

• You don't have a close circuit, there is no path for current to flow around. Commented Mar 19, 2016 at 11:44
• Yeah, but I want to know it in respect of movement of electrons.. Commented Mar 19, 2016 at 11:55
• Current is the movement of electrons. Commented Mar 19, 2016 at 11:58
• 'Voltage' is not the movement of electrons. Commented Mar 19, 2016 at 13:07

At the time of switch-on the charge on the capacitor is zero and the voltage across it is zero.

After switch-on the charge on the capacitor is still zero (and the voltage across it is still zero). At this point, however, the left side of the capacitor has risen to 1 V so the right side must also.

The situation changes if you add a resistor to the Vout terminals. The right side of the capacitor will jump up with the left side but will discharge to 0 V at a rate determined by R x C. The final state will be 1 V across the capacitor.

• I just want to know why right side of capacitor 'must' rise to 1v, not because 'just charge on the capacitor is still zero' but something like, electrons coming out of right side of capacitors.. etc.. Commented Mar 19, 2016 at 12:17
• OK. "Right side of capacitor must rise to 1 V because no electrons flow in or out of it." Commented Mar 19, 2016 at 12:19

Current measures the movement of electrons

Voltage measure the electron's potential energy, or in laymans terms, their desire to move.

A high voltage means the electrons will move a lot if they can, so a high voltage means a high current if they can move. In this case, the output of the capacitor is floating, so they can't move. But they can still have a voltage, because the voltage on the other side of the capacitor is pushing them out.

All the voltage measurements are differential. That is we measure Vin+ to Vin-, Vout+ to Vout-.

If you prefer, you can look at it this way.

As no current flows round the circuit, the voltage across the capacitor does not change, it stays at zero. The Vin graph you posted shows the Vin- voltage going 1V 'down' with respect to the Vin+ terminal. As the voltage across the capacitor stays at 0V, the Vout- terminal also goes 1V down with respect to the Vout+ terminal, as shown in the Vout graph. There's no need to worry about the voltages on the capacitor terminals 'moving' at all.

Of course, as there's no other voltage reference drawn, no ground terminal, it is only differential voltages that are meaningful. So the question of do the capacitor terminals go up in voltage is meaningless, their voltage can only be referenced to other terminals in the circuit.