# Charging a capacitor with a switch

If you have the positive terminal of a battery connect to a switch then to one plate of a parallel plate capacitor and the negative terminal connected to the other plate, what happens when the switch is still open? Does charge build up on the bottom plate of the capacitor only or does nothing happen? There is an electric potential between the negative terminal and one of the plates so charge should still flow in my opinion even if there is a switch on the opposite side of the circuit. Could someone plz clarify this for me?

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i think charge can flow upto the point where the switch is connected. After that there is no conducting path for flowing of the charge. But Imn't sure of it.

berkeman
Mentor
If you have the positive terminal of a battery connect to a switch then to one plate of a parallel plate capacitor and the negative terminal connected to the other plate, what happens when the switch is still open? Does charge build up on the bottom plate of the capacitor only or does nothing happen? There is an electric potential between the negative terminal and one of the plates so charge should still flow in my opinion even if there is a switch on the opposite side of the circuit. Could someone plz clarify this for me?
Welcome to the PF, Seiken. I moved your question to the Homework Help forums, where we have homework and coursework type questions.

In answer to your question, Q=CV still, even with the switch open, but the C capacitance is the capacitance across the open switch, in series with the real capacitor. And capacitances in series add through the inverse formula, so the small capacitance across the open switch dominates. That means that there is very little charge stored by the voltage source, since it is just storing charge across the switch's very small capacitance.

Make sense?