Finding the capcitance of this capacitator

[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

ε_r=ax+b
ε_r=ε₁=a*0+b
b=ε₁
ε_r=ε₂=a*d+ε₁
a=(ε₂-ε₁)/d

ε_r=(ε₂-ε₁)x/d+ε₁
ε_r=((ε₂-ε₁)x+ε₁d)/d

know this capacitator C1 is like millions of tiny little capacitators, dC, each with dialectric substance changing like ε_r=((ε₂-ε₁)x+ε₁d)/d, all connected in a column,

1/C_p=\intd/(Aε₀ε_r)d(ε_r) from (ε₁ to ε₂)

=d/(Aε₀)\intd(d/((ε₂-ε₁)x+ε₁d))

=d²/(Aε₀)\intdx/((ε₂-ε₁)x+ε₁d) (from 0 to d)

=d²/(Aε₀)*1/(ε₂-ε₁)* ln((ε₂-ε₁)x+ε₁d)|from 0 to d

=[d²ln(ε₂/ε₁)]/ε₀(ε₂-ε₁)Aε₀


but the correct answer is meant to be 1/C_p=[dln(ε₂/ε₁)]/ε₀(ε₂-ε₁)Aε₀

can anyone see whre i went wrong?

 

[Redbelly98]

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

ε_r=ax+b
ε_r=ε₁=a*0+b
b=ε₁
ε_r=ε₂=a*d+ε₁
a=(ε₂-ε₁)/d

ε_r=(ε₂-ε₁)x/d+ε₁
ε_r=((ε₂-ε₁)x+ε₁d)/d

know this capacitator C1 is like millions of tiny little capacitators, dC, each with dialectric substance changing like ε_r=((ε₂-ε₁)x+ε₁d)/d, all connected in a column,

1/C_p=∫d/(Aε₀ε_r)d(ε_r) from (ε₁ to ε₂)

=d/(Aε₀)∫d(d/((ε₂-ε₁)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/((ε₂-ε₁)x+ε₁d) (from 0 to d)

=d²/(Aε₀)*1/(ε₂-ε₁)* ln((ε₂-ε₁)x+ε₁d)|from 0 to d

=[d²ln(ε₂/ε₁)]/ε₀(ε₂-ε₁)Aε₀


but the correct answer is meant to be 1/C_p=[dln(ε₂/ε₁)]/ε₀(ε₂-ε₁)Aε₀

can anyone see whre i went wrong?

 

[Dell]

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

 

[Redbelly98]

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(ε₂/ε₁)] / [ε₀ (ε₂-ε₁) A ε₀],​

because the units are inconsistent.

 

[Dell]

and the units of my answer??

 

[Redbelly98]

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

 

[Redbelly98]

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.

 

[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

 

[Redbelly98]

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

 

[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

 

[Redbelly98]

e0 does have units, e1 and e2 do not.

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

 

[Dell]

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

1/C=dx/(ε₀ε_rA)
=dx/(ε₀((ε2-ε1)x+ε1D)/D)A)

i used D for the distance

= D/(ε₀A)∫dx/((ε₂-ε₁)x+ε₁D)
=D/(ε₀A(ε₂-ε₁)) * [ln(ε₂/ε₁)]
-----------------------------------------------------------------------------------

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

 

[Redbelly98]

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.

 

[Redbelly98]

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/(ε₀ε_rA)
=dx/(ε₀((ε2-ε1)x+ε1D)/D)A)

i used D for the distance

= D/(ε₀A)∫dx/((ε₂-ε₁)x+ε₁D)
=D/(ε₀A(ε₂-ε₁)) * [ln(ε₂/ε₁)]