Change in Voltage across a Capacitor?

[Dekans6]

When you increase the charge Q on a capacitor, why don't you increase voltage V across it as well? Wouldn't higher Q increase E, and since V = Ed for a uniform E, V increases also? I know that for a uniform E across a capacitor, E stays the same regardless of distance d between the two plates, but doesn't the amount of charge on the capacitor still affect E (and would therefore affect V)?

 

[Fenn]

I would agree with you, that an increase in Q should increase V. The capacitance of a capacitor is given by

<br /> C = \frac{Q}{V}<br />

where C depends on the physical properties of the capacitor. Stuff like the plate geometry and the medium between the plates. If C is constant, then V should scale linearly with Q.

 

[Dadface]

I would agree with you, that an increase in Q should increase V. The capacitance of a capacitor is given by

<br /> C = \frac{Q}{V}<br />

where C depends on the physical properties of the capacitor. Stuff like the plate geometry and the medium between the plates. If C is constant, then V should scale linearly with Q.

true Q=CV

 

[Dekans6]

Also, just to clarify:

The difference b/w a charged capacitor disconnected from a battery and one that is still connected to a battery is that the latter maintains a constant V across the capacitor regardless of distance d between the plates, and the former maintains a constant Q on the plates (assuming there's no discharge)?


V = Ed

C = eA/d

Q = CV

Thanks

 

[Fenn]

That is correct. If you are adjusting the capacitance by any method, then the voltage will remain constant if there is an applied potential from a battery. If the capacitor is disconnected from any external device, the charge cannot flow, and thus Q will remain constant.

 

[Dadface]

That is correct. If you are adjusting the capacitance by any method, then the voltage will remain constant if there is an applied potential from a battery. If the capacitor is disconnected from any external device, the charge cannot flow, and thus Q will remain constant.

And in addition to this if C is adjusted with the battery connected it takes time for V to come back to its original value this time depending on the time constant(RC) of the circuit.I think this is the principle behind the capacitor microphone.