- #1

jezza10181

- 13

- 1

- Homework Statement
- Capacitor/Resistor network Question

- Relevant Equations
- Q = V * C

Hello,

I have a question regarding the capacitor/resistor network as shown.

My question is simple. I realize that the instant the switch is closed, then the top plate of the capacitor must be at a potential (VA) of 5v.

However, I also realize that the instant the switch is closed, literally at time zero, then the capacitor must have zero charge on it. In that case, then the bottom plate must also be at a potential (VB) of 5v.

This also makes sense because for current to flow through the resistor, then there needs to be a potential difference across it. My question is this, how does the bottom plate of the capacitor acquire a potential of 5v? I don't understand how this happens on a 'microscopic' level. How does it happen?

Regards

Jeremy Watts

I have a question regarding the capacitor/resistor network as shown.

My question is simple. I realize that the instant the switch is closed, then the top plate of the capacitor must be at a potential (VA) of 5v.

However, I also realize that the instant the switch is closed, literally at time zero, then the capacitor must have zero charge on it. In that case, then the bottom plate must also be at a potential (VB) of 5v.

This also makes sense because for current to flow through the resistor, then there needs to be a potential difference across it. My question is this, how does the bottom plate of the capacitor acquire a potential of 5v? I don't understand how this happens on a 'microscopic' level. How does it happen?

Regards

Jeremy Watts