# Homework Help: Finding the capcitance of this capacitator

1. Apr 24, 2009

### Dell

given the capacitator C1 filled with dialectric subtances, which change linearly from ε1 to ε2, what is the capacitance of C1?

since i know that ε changes linerly, according to the distance between the plates, and i know that at x=0 ε=ε1 and x=d ε=ε2

εr=ax+b
εr1=a*0+b
b=ε1
εr2=a*d+ε1
a=(ε21)/d

εr=(ε21)x/d+ε1
εr=((ε21)x+ε1d)/d

know this capacitator C1 is like millions of tiny little capacitators, dC, each with dialectric substance changing like εr=((ε21)x+ε1d)/d, all connected in a column,

1/Cp=$$\int$$d/(Aε0εr)d(εr) from (ε1 to ε2)

=d/(Aε0)$$\int$$d(d/((ε21)x+ε1d))

=d2/(Aε0)$$\int$$dx/((ε21)x+ε1d) (from 0 to d)

=d2/(Aε0)*1/(ε21)* ln((ε21)x+ε1d)|from 0 to d

=[d2ln(ε21)]/ε021)Aε0

but the correct answer is meant to be 1/Cp=[dln(ε21)]/ε021)Aε0

can anyone see whre i went wrong?

2. Apr 25, 2009

### Redbelly98

Staff Emeritus
While I don't exactly follow what you are doing, it appears that your extra factor of d first appears here.

Just as a general rule, I wouldn't name a variable d if integration will be involved in solving the problem. Too easy to lose track of which is the variable d, and which is used to denote the variable of integration.

3. Apr 25, 2009

### Dell

thats exactly right, the problem comes from the fact that i already have one 'd' and my εr is also dependant on 'd' εr=((ε2-ε1)x+ε1d)/d and thats where the second factor comes from, but i cannot see why this should be incorrect mathematically.

4. Apr 25, 2009

### Redbelly98

Staff Emeritus
I would have set up the integral differently to begin with. What is 1/C for a slab of thickness dx? (d→differential, not capacitor thickness here) Then add them by doing the appropriate integral, as you are aware. It's not clear to me how you got the first integral you wrote.

p.s. there is a problem with the "correct" answer you quoted,

1/Cp=[d ln(ε21)] / [ε021) A ε0],​

because the units are inconsistent.

5. Apr 25, 2009

### Dell

and the units of my answer??

6. Apr 25, 2009

### Redbelly98

Staff Emeritus
It would be beneficial for you to work that out. They should come out to be 1/F.

7. Apr 25, 2009

### Redbelly98

Staff Emeritus
I'll offer this hint:

The quantity d / (A ε0) has the correct units of 1/Farads. So any remaining terms should, overall, be unitless.

8. Apr 25, 2009

### Dell

d / (A ε0) = 1/F

d2ln(ε2/ε1)]/(ε2-ε1)Aε0=d/(A ε0)* (d*ln(ε2/ε1)/(ε2/ε1))

what would the units of (d*ln(ε2/ε1)/(ε2/ε1)) be

9. Apr 25, 2009

### Redbelly98

Staff Emeritus
The ε's are all unitless, leaving ... ?

10. Apr 25, 2009

### Dell

so i get m/F,

but you said that there was a problem with the "correct " answers units, as far as i see they come to 1/F

1/Cp=[dln(ε2/ε1)]/(ε2-ε1)Aε0

i just saw that i wrote
1/Cp=[dln(ε2/ε1)]/ε0(ε2-ε1)Aε0
and there is an extra ε0, but you said that they are unitless

11. Apr 25, 2009

### Redbelly98

Staff Emeritus
e0 does have units, e1 and e2 do not.

So, the "correct" answer has the correct units afterall.

Last edited: Apr 25, 2009
12. Apr 26, 2009

### Dell

so how would you have solved it?
the capacitance you spoke of would be

1/C=dx/(ε0εrA)
=dx/(ε0((ε2-ε1)x+ε1D)/D)A)

i used D for the distance

= D/(ε0A)∫dx/((ε21)x+ε1D)
=D/(ε0A(ε21)) * [ln(ε21)]
-----------------------------------------------------------------------------------

if i would not have taken 1/C but instead C, why would i not reach the correct answer,

13. Apr 26, 2009

### Redbelly98

Staff Emeritus
The slabs of thickness dx act as many capacitors in series.

For capacitors in series, we add the reciprocals of capacitance to get total capacitance.

14. Apr 26, 2009

### Redbelly98

Staff Emeritus
Oh, I forgot to say in my last post ... I would do exactly what you did here: