# Parallel plate capacitor capcitance question?

1. Jul 6, 2006

### mazinse

I know that the capacitance is depended on area and distance etc etc. However, can someone explain to me the statement that as the area increases the, Elctrical field decreases.
C would increase because of the area, and V would decrease because of the E, so for this scenario would the charge be constant or increasing.

Can someone show in equation forms, please show how E would decrease.

maybe a little concept too. This is something I cant understand.

2. Jul 6, 2006

### nazzard

Hello mazinse,

The formula to determine the capacity of a parallel plate capacitor (distance of the plates being d) would be:

$$C=\epsilon_0\epsilon_r*\frac{A}{d}$$

As you said with increasing area A the capacity increases as well.

Now the formula for the electric field:

$$E=\frac{U}{d}$$

$$U=\frac{Q}{C}=\frac{Q}{\epsilon_0\epsilon_r*\frac{A}{d}}$$

$$E=\frac{Q}{\epsilon_0\epsilon_rA}$$

As the area increases the electic field decreases.

In your scenario only the area is increased, the charge would remain the same.

Regards,

nazzard

Edit: Aaaahhhh, the forum won't stop eating my equations up

Last edited: Jul 6, 2006
3. Jul 6, 2006

### mazinse

yes it's wonderful using the energy equation, this works out nice. However, why does electrical field decrease, can u give me a conceptual explanation.

4. Jul 6, 2006

### nrqed

Notice that all the "U"s in nazzard's post are actually potential differences $\Delta V$, not the usual energy density U (his notation is not conventional).

To answer your question, the total E field at a point is ismply the vector sum of the E fields produced by all the charges, right? Now imagine uniformly spreading the same amount of charges over a larger area. Then of course the total E field will be weaker (since almost all the charges will be farther away from the point).

5. Jul 8, 2006

### mazinse

ok ok that is helpful, but why is charge constant when A goes up, what controls it? If I increase like the current or voltage of the power supply, would the charge change then? But that would change potential difference too then right? And everything would be changed

6. Jul 8, 2006

### arunbg

It is the change in potential difference applied that causes the change in charge.
If charge were to increase by m times and area increased by n times the change in the electric fieold would be m/n times the original, as can be seen by using the relation posted by nazzard,
$$E=\frac{Q}{\epsilon_0\epsilon_rA}$$