• Support PF! Buy your school textbooks, materials and every day products Here!

Finding the capcitance of this capacitator

  • Thread starter Dell
  • Start date
  • #1
590
0
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=[tex]\int[/tex]d/(Aε0εr)d(εr) from (ε1 to ε2)

=d/(Aε0)[tex]\int[/tex]d(d/((ε21)x+ε1d))

=d2/(Aε0)[tex]\int[/tex]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?
 

Answers and Replies

  • #2
Redbelly98
Staff Emeritus
Science Advisor
Homework Helper
Insights Author
12,098
129
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=d/(Aε0εr)d(εr) from (ε1 to ε2)

=d/(Aε0)d(d/((ε21)x+ε1d))
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.

=d2/(Aε0)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?
 
  • #3
590
0
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
Redbelly98
Staff Emeritus
Science Advisor
Homework Helper
Insights Author
12,098
129
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
590
0
and the units of my answer??
 
  • #6
Redbelly98
Staff Emeritus
Science Advisor
Homework Helper
Insights Author
12,098
129
It would be beneficial for you to work that out. They should come out to be 1/F.
 
  • #7
Redbelly98
Staff Emeritus
Science Advisor
Homework Helper
Insights Author
12,098
129
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
590
0
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
Redbelly98
Staff Emeritus
Science Advisor
Homework Helper
Insights Author
12,098
129
The ε's are all unitless, leaving ... ?
 
  • #10
590
0
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
Redbelly98
Staff Emeritus
Science Advisor
Homework Helper
Insights Author
12,098
129
e0 does have units, e1 and e2 do not.

So, the "correct" answer has the correct units afterall.
 
Last edited:
  • #12
590
0
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
Redbelly98
Staff Emeritus
Science Advisor
Homework Helper
Insights Author
12,098
129
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
Redbelly98
Staff Emeritus
Science Advisor
Homework Helper
Insights Author
12,098
129
so how would you have solved it?
Oh, I forgot to say in my last post ... I would do exactly what you did here:
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)]
 

Related Threads for: Finding the capcitance of this capacitator

Replies
5
Views
5K
  • Last Post
Replies
5
Views
1K
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
4
Views
2K
  • Last Post
Replies
3
Views
3K
  • Last Post
Replies
1
Views
737
  • Last Post
2
Replies
31
Views
5K
Top