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Parallel plate capacitor capcitance question? 
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#1
Jul606, 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. 


#2
Jul606, 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 


#3
Jul606, 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.



#4
Jul606, 11:36 PM

Sci Advisor
HW Helper
P: 2,887

Parallel plate capacitor capcitance question?
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
Jul806, 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



#6
Jul806, 02:17 PM

P: 601

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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