- #1
jmachu2
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Hello. Firstly, I'm new to these forums, so if this topic belongs elsewhere, I apologize.
I've been struggling with certain concepts pertaining to electrostatics, and would like some feedback regarding as to whether or not I have the right idea in understanding how charge passes through a capacitor.
For example, let's say we have a simple electrical circuit (i.e. our circuit only has a battery source and a parallel plate capacitor). When we turn on our battery, we create current (I) that flows from the + battery source to the capacitor plate (C) through a copper wire; because our copper wire is composed of atoms, when we turn on the battery, creating an electric field (E) through the wire, the electrons within the wire will shift towards the + battery source. Thus, due to a deficiency in electrons in the wire near (C), our capacitor plate nearest to the + battery will gain a + charge (q+). The capacitor plate will continue to gain q+ until the positive charges build up on the plate, and eventually are repulsed by each other. When this happens, (I) will stop flowing and (C) will be fully charged. The same thing happens with the - battery source as well, though opposite (i.e. the electrons will flow to the capacitor plate, instead of away from it.)
(At this point, if you see anything that's wrong please offer some input.)
Now, onto my main question: How does charge pass through a capacitor?
Let's say between our capacitor plates, we have air, which is a dielectric. When our capacitor plate is fully charged, the atoms within the dielectric will polarize to align their appropriate charge with that of the proper capacitor plate; the electrons in the atoms of the dielectric will shift towards the + plate, and vice versa. I understand that it is because of this effect that our capacitor can now contain even more charge on it from the relationship C = q/Voltage. However, my question arises as to how the charges propagate from one plate to the other when the capacitor discharges. Does the charge simply pass through the dielectric material onto the opposite plate?
Any help will be appreciated. Thank you.
I've been struggling with certain concepts pertaining to electrostatics, and would like some feedback regarding as to whether or not I have the right idea in understanding how charge passes through a capacitor.
For example, let's say we have a simple electrical circuit (i.e. our circuit only has a battery source and a parallel plate capacitor). When we turn on our battery, we create current (I) that flows from the + battery source to the capacitor plate (C) through a copper wire; because our copper wire is composed of atoms, when we turn on the battery, creating an electric field (E) through the wire, the electrons within the wire will shift towards the + battery source. Thus, due to a deficiency in electrons in the wire near (C), our capacitor plate nearest to the + battery will gain a + charge (q+). The capacitor plate will continue to gain q+ until the positive charges build up on the plate, and eventually are repulsed by each other. When this happens, (I) will stop flowing and (C) will be fully charged. The same thing happens with the - battery source as well, though opposite (i.e. the electrons will flow to the capacitor plate, instead of away from it.)
(At this point, if you see anything that's wrong please offer some input.)
Now, onto my main question: How does charge pass through a capacitor?
Let's say between our capacitor plates, we have air, which is a dielectric. When our capacitor plate is fully charged, the atoms within the dielectric will polarize to align their appropriate charge with that of the proper capacitor plate; the electrons in the atoms of the dielectric will shift towards the + plate, and vice versa. I understand that it is because of this effect that our capacitor can now contain even more charge on it from the relationship C = q/Voltage. However, my question arises as to how the charges propagate from one plate to the other when the capacitor discharges. Does the charge simply pass through the dielectric material onto the opposite plate?
Any help will be appreciated. Thank you.