Parallel plate capacitor capcitance question?

[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 can't understand.

 

[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

 

[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.

 

[nrqed]

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.

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).

 

[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

 

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