# introductory physics concept

posted by on .

I know that a capacitor does not allow charge to flow through it because its two conducting plates are separated by a dielectric (insulating material) but then how does a current flow through a circuit? Is it because at a certain DC voltage (almost like a breaking point), the insulating material in between the conducting plates breaks down and the voltage is able to pass through? I read this online in a scientific forum, but I just wanted to clarify that this is correct. Can someone please provide me with a thorough explanation? I want to make sure I have my basics down correctly. Thank you everyone for your help.

• introductory physics concept - ,

Can someone also tell me if this amount of voltage varies from one circuit to another? If so, how would you know? Is this an explanation for the bulb not lighting up in a simple circuit where the capacitor is after the switch (which is also right after the battery), but before the bulbs? Thank you very much.

• introductory physics concept - ,

The charges do not have to flow "through" the capacitor. They merely flow into one side of the capacitor, building up the voltage as long as they flow and build up. If the voltage on the capacitor is not changing, then no current is flowing into or out of it. The capacitor is a storage device for charges, sort of like having a tank of water on your roof. As long as you pump water up to the tank, there is a flow, and the water gets higher in the tank increasing the potential energy. Just like the capacitor you can then get flow out of the tank later after turning the pump off.
If you turn the switch off, there is no flow and the bulb can not light. That is just like turning a valve off in the pipe to that tank in your attic. You still have voltage, but no current.

• introductory physics concept - ,

But I still do not understand how the bulbs will light up if the charge is being stored on one side only ,and the bulb is closer to the other plate on the other side of the capacitor?