Capacitors and Dielectric Oil: Calculating Electric Displacement and Field

[teme92]

Homework Statement

(a) Calculate the Electric displacement of parallel plate capacitor of with a thickness d (in z), and length L (in x), and a width W (in y). Within the capacitor is a dielectric oil with a dielectric constant –ε_r , and the charge on the plates is ± q.
(b) Using the definition of Electric displacement, what is the Electric field within the capacitor?
(c) Calculate the capacitance? (
d) If the capacitor is turned sideways so that the width W is vertical, while the length L and thickness d are horizontal, how much oil will remain within the capacitor? (provide either a volume or a length) Assume that the density of the dielectric oil is ρ . (Warning: your final solution may not be closed form and if so, need not be fully solved)

Homework Equations

The Attempt at a Solution

(a) Used Gauss' Law:

$\int D{\cdot}da = Q$ where $\int D{\cdot}da = 2DA$.

Therefore: $D = \frac{Q}{2A} = \frac{\sigma}{2}\hat{z}$

(b) D is proportional to electric field as dielectric material is linear:

$E = -\frac{\sigma}{2\epsilon_r}$

(c) Used the formula:

$C = \epsilon_r\epsilon_0\frac{A}{d}$

(d) Have no idea how to begin this part even. Any help in the right direction would be greatly appreciated.

 

[mfb]

Therefore: $D = \frac{Q}{2A} = \frac{\sigma}{2}\hat{z}$

The middle part is a scalar, the right side is a vector, those cannot be equal.
I don't think it helps to introduce new variables here, especially if you don't give a definition of them.

(d): Oil will move down until the height reaches some value h. At this height, removing more oil from the bottom does not release energy any more as moving oil out of the electric field takes energy, so no more oil moves out. You can calculate this height h.
The problem statement could be a bit clearer here - if you would try this with a real capacitor various other effects will lead to completely different results. You have to put the capacitor on top of an oil bath to get the same result in a real experiment.

 

[teme92]

Ok thanks for the help, I'm working through and I'll get back to you if I need any more help.