Homework Help: Capacitors and Dialectrics

1. Mar 4, 2012

forestmine

EDIT: That ought to be, "Dielectrics," of course :)

1. The problem statement, all variables and given/known data

An air capacitor is made by using two flat plates, each with area A, separated by a distance d. Then a metal slab having thickness a (less than d) and the same shape and size as the plates is inserted between them, parallel to the plates, and not touching either plate.

a) What is the capacitance of this arrangement?
b) Express the capacitance as a multiple of the capacitance C_0 when the metal slab is not present.

2. Relevant equations

C = KC_0
C_0 = ε_0A/d

3. The attempt at a solution

At first glance, I thought this should be a pretty quick and painless process. If the capacitance of capacitors with a dielectric is C = KC_0, and C_0 in this case is ε_0A/(d-a)/2, then C=(Kε_0A)/[(d-a)/2] but the correct answer turns out to be ε_0A/(d-a) and I'm not really sure why that is?

For part b, I would have thought that the answer is simply C = ε_0A/d, but again that is incorrect.

Seems like I'm just missing some simple relation. If anyone could lend a hand, I'd really appreciate it!

Thank you :)

Last edited: Mar 4, 2012
2. Mar 4, 2012

Staff: Mentor

Inserting a metal plate between the two plates of a capacitor effectively divides the capacitor into two in series (each face of the metal slab acts as a new capacitor plate for the original capacitor plate that it faces). So you get two capacitors in series, each has the same plate area as the original but a decreased plate separation.

3. Mar 4, 2012

forestmine

That makes perfect sense, and makes things extremely easy from there.

Thank you so much again!