Question on Capacitors

[TawMus]

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The electric field between 2 charged capacitors is always pointing away from the + plate and into the - plate. During the charging process, the electrons from the - plate are transferred onto the + plate against the electrical field w/ work being done by an external force(Voltage).

So here's my question, while the transfer of electrons is occuring... once the charges on each plate get past a threshold, do the plate charges and electric field invert in the opposite direction? (+ changes to - plate, and vice versa... and E field switches too since it must always point towards the negative plate)

And while the capacitor is discharging, do the plate charges (+Q, -Q) also swap (+Q becomes -Q, -Q becomes +Q) as the electrons return to their normal plates??

Is this reasoning right? Or am I missing something? Thanks.

 

[BerkReviewTeach]

The electric field between 2 charged capacitors is always pointing away from the + plate and into the - plate. During the charging process, the electrons from the - plate are transferred onto the + plate against the electrical field w/ work being done by an external force(Voltage).

The bolded part above is backwards, which might be the start of the problem. The charging process does in fact require work, but that is because the charges are buidling up on the respective plates. Over time the accumulated charge causes enough repulsion to electron flow that it offsets the voltage and the plates are said to be charged. This occurs because electrons (negative charges) continue to flow from the positive plate to the negative plate (which is why work is required).

So here's my question, while the transfer of electrons is occuring... once the charges on each plate get past a threshold, do the plate charges and electric field invert in the opposite direction? (+ changes to - plate, and vice versa... and E field switches too since it must always point towards the negative plate)

Assuming to are talking about a DC circuit, nothing reverses until you connect the capacitor to a different circuit. If we consider just the charging process, you are simply filling up the plates with their respective charges until the push of the voltage is offset by the repulsion of the charged plates to further charges. At this time, the plates reach a maximum charge which can be determined using Q = VC.

And while the capacitor is discharging, do the plate charges (+Q, -Q) also swap (+Q becomes -Q, -Q becomes +Q) as the electrons return to their normal plates?

If it's a DC circuit, the charges simply come off of the plates until the plates are once again neutral. Over the course of time, the charge magnitude drops, so the repulsion that pushes charge off of the capacitor plates decreases, and eventually drops to zero. However, if you are talking about an AC circuit, then it would in fact refill in the opposite manner until the charges are reversed as you described (assuming the fill time is less than period of the circuit OR the AC follows a perfect sine function).

 

[RySerr21]

i have another capacitor question....

okay so if you add a dielecric to a capacitor, that will increase the capicatance because it decreases the voltage, right? Does anything happen to the charge??

What if you add a dielectric but the question stem says it is hooked up to a constant voltage? then does the charge increase?

 

[TawMus]

i have another capacitor question....

okay so if you add a dielecric to a capacitor, that will increase the capicatance because it decreases the voltage, right? Does anything happen to the charge??

What if you add a dielectric but the question stem says it is hooked up to a constant voltage? then does the charge increase?

Yes, adding a dielectric to a capacitor ALWAYS increases capacitance. Voltage is decreased if there is no voltage source and charge increases if there is a voltage source.

 

[unsung]

Yes, adding a dielectric to a capacitor ALWAYS increases capacitance. Voltage is decreased if there is no voltage source and charge increases if there is a voltage source.

Yep, I think of it as the dielectric prevents charge from "seeping" through into the space between the plates, and in that way allows for a greater charge to be observed.

Since C = Q/V, increased Q means increased capacitance C. (This is for if a voltage source exists, of course.)