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Parallel plate capacitor capcitance question?

by mazinse
Tags: capacitor, capcitance, parallel, plate
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mazinse
#1
Jul6-06, 02:13 PM
P: 190
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.
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nazzard
#2
Jul6-06, 02:22 PM
PF Gold
P: 100
Hello mazinse,

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

[tex]C=\epsilon_0\epsilon_r*\frac{A}{d}[/tex]

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

Now the formula for the electric field:

[tex]E=\frac{U}{d}[/tex]

[tex]U=\frac{Q}{C}=\frac{Q}{\epsilon_0\epsilon_r*\frac{A}{d}}[/tex]

[tex]E=\frac{Q}{\epsilon_0\epsilon_rA}[/tex]

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
#3
Jul6-06, 11:20 PM
P: 190
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
#4
Jul6-06, 11:36 PM
Sci Advisor
HW Helper
P: 2,948
Parallel plate capacitor capcitance question?

Quote Quote by 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.
Notice that all the "U"s in nazzard's post are actually potential differences [itex] \Delta V [/itex], 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
#5
Jul8-06, 01:56 PM
P: 190
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
#6
Jul8-06, 02:17 PM
P: 600
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,
[tex]E=\frac{Q}{\epsilon_0\epsilon_rA}[/tex]


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