# Capacitance 3 dielectrics sandwiched between two conductors.

1. Oct 8, 2013

### MostlyHarmless

Find that capacitance of the system consisting of 3 dielectrics w/ length, l=1.0m, width, w= 1.0m and depth, d= 1.0 CM. $k_1=1.5, k_2=2, k_3=2.5$ *dimensions of conducting plates not given*

Equations: Capacitance, $C= Q/{\delta}V$

Field in the dielectrics
$E={\frac{\sigma}{k{\epsilon}_0}}$

I've found the fields in each dielectric, and the potential difference in each, but I'm getting hung up there. I've gotten down to, $|{\delta}V|=({\frac{Q}{A{\epsilon}_0}})({\frac{1}{k_1}}+{\frac{1}{k_2}}+{\frac{1}{k_3}})$

My problem is, the A refers to the area of the conducting plate, but as i pointed out, nothing is said concerning dimensions of those plates.

Is there another method that does not care about the area that I'm missing?

2. Oct 8, 2013

### Staff: Mentor

You can assume the plates have the same dimensions as width and length of the dielectric materials.

3. Oct 8, 2013

### MostlyHarmless

That's what I thought, but the diagram provided clearly shows that conductors being larger than the dielectrics.

Is the portion that is the same dimension as the dielectrics the only portion if the conductor that has any effect?

4. Oct 8, 2013

### Staff: Mentor

Can you show this diagram?
No. For large k this is a good approximation (as you can neglect the air then), but your k-values are not large.