- #1

Garoll

- 8

- 0

I can`t understand the following:

Let`s say that we have a simple capacitor with two plates and air between them. And this capacitor is in a series circuit with a switch, voltage source and a resistor. Also, the capacitor is discharged and the switch is open. The voltage source is 10V for example, DC. The circuit is 1. Voltage source, 2. Switch, 3. Capacitor, 4. Resistor.

Now we are closing the switch. It is said that at the instant of closing the switch there is no charge on the on the capacitor and therefore no potential difference across it. As a result the whole of the applied voltage must momentarily be applied on the resistor.

I can`t understand how these 10 positive volts will pass through the air of the capacitor. At the instant time of switching on the left on the capacitor there will be 10V and there will be no potential difference, therefore on the right of the capacitor there will be 10V too. I can`t understand how at the instant moment this positive charge will pass through the capacitor and the voltage will be 10 Volts on both plates.

I can see on the oscilloscope that this is true, for example when having input of square impulses on a differentiator circuit it is exactly like i explained - there is a pick value on the output when the impulse just starts.

I can`t understand how physically this positive charge appears on the right side of the capacitor.

I would be happy if someone can explain this.

Best regards