Question

Is there anyone here that can help me with electrical capasitors? Involving permittivity, plates, etc.

Answer 1

Roughly translated from norwegian:

"The insulation in a plate-capasitor consists of two layers with a thickness of 1 and 2 mm, and a relative permittivity of 4 and 6, respectively. Calculate the electric field strength in the thinnest layer when the capasitor is connected to a voltage source of 5000 V."

Answer 2

The easiest, way to think of this is as two capacitors in series, one has a plate separation of 1 mm and is filled with insulator 1, and the second has plate separation 2 mm and insulator 2.

Answer 3

I've been ripping my hair trying to get this! :P but how do I do this without knowing the area of the plates?

Answer 4

You can find the ratios of the capacitances, which doesn't depend on the area.
Then you can find the ratio of the voltages.
Since the voltages add to 5000 V, you can get the individual voltages

Answer 5

But, I can't find a formula that calculates the capacitance without using the area! :-/

Answer 6

You know for a parallel plate capacitor \[C=\epsilon A/d\], so \[C_1/C_2 ={{\epsilon_1 d_2}\over {\epsilon_2 d_1}}\].

You also know that the charge on capacitors in series is the same, so that
\[Q= C_1V_1=C_2V_2\]

and that the voltages add so \[V_1 + V_2 = 5000 V\]

The electric field inside a parallel plate cap[acitor is uniform, so
\[ V_1= E_1 d_1\]

Answer 7

$$
C_1=\cfrac{Q_1}{V_1}=\varepsilon_{r1}\varepsilon_0\frac{A}{d_1}\\
V_1=\cfrac{Q_1d_1}{4\varepsilon_0A}\\
C_2=\cfrac{Q_2}{V_2}=\varepsilon_{r2}\varepsilon_0\frac{A}{d_2}\\
V_2=\cfrac{Q_2d_2}{6\varepsilon_0A}\\
V=5000=V_1+V_2\\
=\cfrac{Q_1d_1}{4\varepsilon_0A}+\cfrac{Q_2d_2}{6\varepsilon_0A}\\
=\cfrac{3Q_1d_1+2Q_2d_2}{12\varepsilon_0A}\\
=\cfrac{Q(3d_1+2d_2)}{12\varepsilon_0A}\\
A=\cfrac{Q(3d_1+2d_2)}{12\varepsilon_0V}\\
d_1=.001~m\\
d_2=.002~m\\
E_1=\cfrac{V_1}{d_1}=\cfrac{Q_1d_1}{4\varepsilon_0Ad_1}=\cfrac{Q_1}{4\varepsilon_0A}=\cfrac{Q}{4\varepsilon_0\cfrac{Q(3d_1+2d_2)}{12\varepsilon_0V}}\\
=\cfrac{3V}{(3d_1+2d_2)}
$$
Note \(Q_1=Q_2=Q\).
Does this make sense?

Answer 8

now thats perfection!

Answer 9

Yes, THANK YOU! :D I just couldn't wrap my head around it, but this helped a lot! :)