Not-So-Parallel Plate Capacitor

[Enemy0fGods]

1. You have fabricated a parallel plate capacitor in your work shop, but the square metal plates end up not being exactly parallel to each other. They form an angle \alpha. The plates, size L, are held at constant electrical potentials, V₁ and V₂, corresponding to an electrical potential difference \DeltaV = V₁ - V₂. Plate 1 holds a +Q charge while plate 2 holds a -Q charge.

I'm done with parts a-c, but I need help with these:

d: Find the total charge carried by the plates (Hint: This requires an integral).
e: Show that the value of the capacitance of this capacitor is C = (\epsilon₀L)/\alpha *ln(b/a). Does this have all the usual attributes of a capacitance?

[PLAIN]http://img19.imageshack.us/img19/220/unledeyr.png

2. Homework Equations : Already listed one, and E=V/d, and the rest I don't know.



3. The Attempt at a Solution : I don't really have an attempt, I'm stuck, all I know is I'm suppose to use an integral with part d, which is already given anyway.

Thanks for any help.

 

[SammyS]

Out of curiosity, what were parts a - c ? & what were your answers to those?

 

[Inna]

2. Homework Equations : Already listed one, and E=V/d, and the rest I don't know.

What do you mean by "d" ? The distance between the plates varies with x.

 

[cupid.callin]

for (e)

consider the part of capacitor inside red rectangle ... it is also a capacitor.
find capacitance of this capacitor in terms of variables you know and x,dx ...
then integrate ...

 

[cupid.callin]

and for d
question itself says plate 1 has charge Q and plate 2 has charge -Q
so total charge would be 0 right?

 

[Idoubt]

and for d
question itself says plate 1 has charge Q and plate 2 has charge -Q
so total charge would be 0 right?

Perhaps the question meant the charge on each plate in terms of potentials and capacitance.

 

[cupid.callin]

but why would question ask that
for (d) first you should do e
so d must be placed next to e ... which it is not