Getting weird formula for Capacitance

[lluke9]

Okay, so I know
$$C=Q/ΔV$$

And ΔV is the sum of the electric fields multiplied by the distance between the charges, so if the first charge has a charge of Q and the other has -Q with R distance between, the electric potential/voltage is:
$$ΔV = [(kQ/r^2) + (kQ/r^2)]R$$
so
$$ΔV = 2kQ/R$$.

And $$C = Q/ΔV$$
so...
$$C = Q/(kQ/R)$$
and...
$$C = R/2k$$
and...
$$C = R/2[1/(4Πε_0)]$$
and
$$C = 2ΠRε_0$$

What...?
It would make some semblance of sense if R were inversely proportional to capacitance, but it isnt...

 

[Doc Al]

Okay, so I know
$$C=Q/ΔV$$

Where ΔV is the voltage between two conductors.

And ΔV is the sum of the electric fields multiplied by the distance between the charges,

That's only true if the field is constant.

so if the first charge has a charge of Q and the other has -Q with R distance between, the electric potential/voltage is:
$$ΔV = [(kQ/r^2) + (kQ/r^2)]R$$

Not sure what you're doing here with two point charges.

In any case: In the expression for the field from a point charge, r is the distance from the charge. So r for one charge is different than the r for the other. Also, the field isn't constant, so you can't just multiply by the distance R.

 

[lluke9]

Where ΔV is the voltage between two conductors.That's only true if the field is constant.Not sure what you're doing here with two point charges.

In any case: In the expression for the field from a point charge, r is the distance from the charge. So r for one charge is different than the r for the other. Also, the field isn't constant, so you can't just multiply by the distance R.

Well, I was just adding up the electric fields and doing V = EΔd...

But I'm guessing that's not possible because the electric field isn't constant...?

Okay, please forget about everything I typed up there, I guess it was a complete waste of time.So how DO you change the voltage in a capacitor? Wouldn't it be to just increase the charge?
Actually, how do you FIND the voltage in a capacitor? Is it V = EΔd, because the electric field is constant in a capacitor?

 

[Doc Al]

So how DO you change the voltage in a capacitor? Wouldn't it be to just increase the charge?

You charge a capacitor by hooking it up to voltage source (a battery, perhaps). The higher the voltage, the greater the charge stored on each conductor.

Actually, how do you FIND the voltage in a capacitor? Is it V = EΔd, because the electric field is constant in a capacitor?

For a parallel plate capacitor, the field is constant. So you could use that method.

 

[lluke9]

For a parallel plate capacitor, the field is constant. So you could use that method.

So could you substitute for voltage in the capacitance equation?

$$C = q/EΔd$$

Then $$E_T$$ would be:
$$E_T = kq/r^2 + kq/r^2$$
because
$$E = kq/r^2$$
and both electric fields are going the same direction.

And then I'd arrive at the same thing I did in my original post...

$$E_T$$ = net electric field

 

[Doc Al]

because
$$E = kq/r^2$$

That's the field for a point charge. Nothing to do with the constant field found between the plates of a parallel plate capacitor.