# Thread: Why is capacitance inversely related to the distance of the conductor plates

1. I need help with something.I have problem to understand how the capacitance of a capacitor can be increased if we move the plates of the capacitor more close.
Here's my reasoning:
The capacitor is just a extended terminal of a battery,so the battery voltage must equal the capacitor voltage.Because the area of the plates of the capacitor is larger than,that of the battery, the plates of the capacitor can accumulate charge until the battery and the capacitor have the same voltage.This is logical for me and I can make a mental picture from it.

I cannot understand how the capacitance can be increase, and voltage not if we move the plates more close.I understand that if we move the plates more close,the positive plate will affect the negative and more charge will accumulate on the negative plate,but if more charge accumulate on the same square are of the plates,that mean that we will have bigger voltage on the capacitor.

For example if we have 1 farad capacitor and the voltage is one volt,we will have one coulomb of charge on the plates.But if we bring the plates more close, then the voltage would be the same but will will have more charge on the same area plates.If we have more charge on a plate with a same area then, the repulsive forces between the electrons should make the voltage bigger.

Thank to anyone who can help me with this.I'm struggling to understand this.

2.

3. Originally Posted by jonah.seler
I need help with something.I have problem to understand how the capacitance of a capacitor can be increased if we move the plates of the capacitor more close.
Here's my reasoning:
The capacitor is just a extended terminal of a battery,so the battery voltage must equal the capacitor voltage.Because the area of the plates of the capacitor is larger than,that of the battery, the plates of the capacitor can accumulate charge until the battery and the capacitor have the same voltage.This is logical for me and I can make a mental picture from it.]/QUOTE]

If your mental picture says that, I regret to inform you that it's wrong. Very wrong.

Whenever you connect open-circuited conductors to battery terminals, the conductors will assume the same potential as the battery. Whether those conductors are called capacitors or not doesn't matter at all. The battery will move charge from one conductor to another until the potentials equal those of the battery. All that happens in the blink of an eye, as they say.

I cannot understand how the capacitance can be increase, and voltage not if we move the plates more close.I understand that if we move the plates more close,the positive plate will affect the negative and more charge will accumulate on the negative plate,but if more charge accumulate on the same square are of the plates,that mean that we will have bigger voltage on the capacitor.
Capacitance is a measure of charge displaced per volt. The closer the plates, the greater the electrostatic forces between them, for a given voltage. That means more charge displaced, and thus more capacitance.

[QUOTE[For example if we have 1 farad capacitor and the voltage is one volt,we will have one coulomb of charge on the plates.But if we bring the plates more close, then the voltage would be the same but will will have more charge on the same area plates.If we have more charge on a plate with a same area then, the repulsive forces between the electrons should make the voltage bigger.

Thank to anyone who can help me with this.I'm struggling to understand this.
You seem to be switching experiments midway. Don't do that.

In your experiment, is the battery disconnected after the initial charging? Or is the battery connected as you move the plates around?

If the former, then the voltage cannot change (there's a battery forcing a constant voltage). Move the plates all you want; the voltage stays equal to the battery voltage.

If the latter, then the displaced charge has nowhere to go. Therefore Q is constant. Now, Q=CV, so the CV product must remain constant. Moving the plates closer together causes the capacitance to go up, which causes the voltage to go down.

The force between the plates is attractive, not repulsive. Maybe that's the source of your confusion?

4. Hi TK421.Thanks for your answer but I still don't understand the problem.

I am considering only the charges of one plate.For example, if we take the negative plate,the negative plate will accumulate charge until the voltage across the capacitor is equal to that of the battery.Now if we bring the plates close together that mean that the negative plate will accumulate more charge on a same square area.{same goes for positive plate,but in a reverse manner}.If we have more electrons on a same square area{or a bigger shortage of electrons on the positive plate} that mean, that the repulsive force between the point charges would be bigger.More electrons,more close together on a same area makes a bigger potential.

5. If we increase the battery voltage,we will increase the capacitance too.If this is correct then,increasing the capacitance by moving the plates more close should increase the voltage.

6. Originally Posted by jonah.seler
If we increase the battery voltage,we will increase the capacitance too.
This is not exact : if you increase the voltage, you will increase the CHARGE (in Coulomb). The capacitance value depends only of the physical characteristics of the device (surface, distance, dielectric). As previously told by TK421, C.V = Q.

Do the following experiment :
- connect a capacitance to a battery. The electric charge of the capacitance will be CV = Q.
- disconnect the battery. Supposing the capacitance is perfect (no electric leak), the charge will remain the same. The voltage will be equal to the one of the battery.
- decrease the distance between the plates : the charge is unchanged but the voltage will decrease.

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