Parallel plate capacitor capcitance question?

by mazinse
Tags: capacitor, capcitance, parallel, plate
 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.
 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: $$C=\epsilon_0\epsilon_r*\frac{A}{d}$$ 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
 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.
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).
 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, $$E=\frac{Q}{\epsilon_0\epsilon_rA}$$