# Parallel plate capacitor problem

1. Feb 27, 2006

### Punchlinegirl

A capacitor is constructed from 2 square metal plates. A dielectric k= 4.93 fills the upper half of the capacitor and a dielectric k=10 fills the lower half of the capacitor. Neglect edge effects. The plate are separated by a distance of 0.37 mm and the length of the plates is 27 cm. Calculate the capacitance C of the device. Answer in units of pF.
I used the equation for C of a parallel plate capacitor.
$$kE_o A/d + kE_o A/d$$
Where the area= 3.7 x 10^-4 * .135 = 4.995 x 10^-5
So (4.93)(8.85 x 10^-12)(4.995 x 10^-5)/ 3.7 x 10^-4 + (10)(8.85 x 10^-12)(4.995 x 10^-5)/ (3.7 x 10^-4)
= 1.78 x 10^-11 F.
Converting to pF gave me 1.78 x 10^-23 which isn't right... help please?

2. Feb 27, 2006

### Astronuc

Staff Emeritus
Last edited: Feb 27, 2006
3. Feb 27, 2006

### Punchlinegirl

I think d has to be different for each one, but I'm not sure how to do that.

4. Feb 27, 2006

### sapta

consider two capacitors in series each having distance d = half of 0.37 mm.The first has dielectric k=4.93,second has k=10.

Or you may use the following equation-

C=EA/[d/k + d'/k'] here d=d'=half of 0.37 mm

5. Feb 27, 2006

### Punchlinegirl

So would I do,
(8.85 x 10^-12)(.27)/ (3.7 x 10^-4/4.93) + ((1.85 x 10^-4)/10) = 2.55 x 10^-8 ?

6. Feb 27, 2006

### sapta

area is (.27 m)^2;also d=1.85 x 10^-4 m.

7. Feb 28, 2006

### Punchlinegirl

Ok so it would be:
(8.85 x 10^-12)(.0729) / ((1.85 x 10^-4/4.93) + (1.85 x 10^-4/10)) =1.15 x 10^-8
Am I doing this right now?

8. Feb 28, 2006

### Astronuc

Staff Emeritus
9. Feb 28, 2006