# Better understanding of capacitor charging

1. ## Re: Better understanding of capacitor charging

Originally Posted by Akanimo
The voltage will not drop. Instead, it will go up.

When you decrease the distance between the plates, the field becomes stronger. This means that the capacitance decreases because you now need less energy to keep the charges on the plates as there's already a stronger attraction between the unlike charges on the two plates.
This also implies that the voltage increases since the charges are still intact. This could be easily seen with this known formula:

Q=CV

Since Q is constant, if C decreases, V must increase proportionately. A lower capacitance requires a lesser amount of charge to increase the voltage by 1V. Remember?
Sorry about that. I wrote it the opposite way. Thanks FvM for pointing that out.

Correction:
The voltage will drop as depicted by formula:

Epsilon*A/d

When you decrease the distance between the plates, the field becomes stronger. This means that the capacitance increases because you now need less energy to keep the charges on the plates as there's already a stronger attraction between the unlike charges on the two plates.
This also implies that the voltage increases since the charges are still intact. This could be easily seen with this known formula:

Q=CV

Since Q is constant, if C decreases, V must increase proportionately. A lower capacitance requires a higher amount of charge to increase the voltage by 1V. Remember?

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2. ## Re: Better understanding of capacitor charging

Originally Posted by Akanimo
Sorry about that. I wrote it the opposite way. Thanks FvM for pointing that out.

Correction:
The voltage will drop as depicted by formula:

Epsilon*A/d

When you decrease the distance between the plates, the field becomes stronger. This means that the capacitance increases because you now need less energy to keep the charges on the plates as there's already a stronger attraction between the unlike charges on the two plates.
This also implies that the voltage increases since the charges are still intact. This could be easily seen with this known formula:

Q=CV

Since Q is constant, if C decreases, V must increase proportionately. A lower capacitance requires a higher amount of charge to increase the voltage by 1V. Remember?
Correction:
The voltage will drop.

When you decrease the distance between the plates, the field becomes stronger. This means that the capacitance increases, as depicted by formula C=Epsilon*A/d, because you now need less energy to keep the charges on the plates as there's already a stronger attraction between the unlike charges on the two plates.
This also implies that the voltage increases since the charges are still intact. This could be easily seen with this known formula:

Q=CV

Since Q is constant, if C decreases, V must increase proportionately. A lower capacitance would have a higher voltage for the same amount of charge difference between the two plates . Remember?

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3. ## Re: Better understanding of capacitor charging

Originally Posted by Akanimo
The voltage will not drop. Instead, it will go up.

When you decrease the distance between the plates, the field becomes stronger. This means that the capacitance decreases because you now need less energy to keep the charges on the plates as there's already a stronger attraction between the unlike charges on the two plates.
This also implies that the voltage increases since the charges are still intact. This could be easily seen with this known formula:

Q=CV

Since Q is constant, if C decreases, V must increase proportionately. A lower capacitance requires a lesser amount of charge to increase the voltage by 1V. Remember?
As you can see from this link http://hyperphysics.phy-astr.gsu.edu...ic/pplate.html , the closer the plates, the higher the capacitance and the lower the voltage. The electrostatic field is stronger when the plates are closer, but it it also shorter.

Ratch

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Originally Posted by c_mitra