estimating potential drops within multimedia capacitorsby Sancor Tags: capacitors, drops, estimating, multimedia, potential 

#1
Nov412, 12:36 PM

P: 7

I was wondering if anyone had any insights on how to estimate or discern the potential drops across a capacitor containing three distinct media between the electrode plates. The system is as follows
Vo(+) air gap  binary electrolyte phase  plastic phase () The lengths of the air gap is approximately that of the electrolyte phase, with the length of the plastic phase being much smaller. I'm trying to figure out a way to consider the the electric field at the interfaces due to the electrical double layer that should be generated by the potential difference across the capacitor. Solving Laplace in one dimension gives I_{1} = Ax+B I_{2} = Cx+D I_{3} = Ex+F Applying boundary conditions I get I_{1}(x=0)=Vo=B I_{3}(x=2L)=0=2EL+F => F = 2EL I_{1}(x=L)=I_{2}(x=L) AL + Vo = CL + D I_{2}(x=2L)=I_{3}(x=2L) 2CL + D = EL  2EL 2CL + D = EL from conservation of charge n*(σ_{1}E_{1}σ_{2}E_{2}) = 0 σ of air and plastic = 0 thus E_{2}=0 C=0 leaving me with 3 unknowns and only two equations AL + Vo = D D = EL Which is why I am trying to see if I can estimate the approximate potential drop due to the air gap, which would allow me eliminate one of my unknowns. Still though, this is only good for determining the potential everywhere within my capacitor, and I am still no closer to determining the electric field due to the double layer at my interfaces. Any ideas? 



#2
Nov512, 12:02 PM

P: 7

So after working on this some more I decided to try and model the system as capacitors in series... and I decided to neglect the plastic layer for this... Anyhow what I ended up doing was to draw my system as a simple circuit with two capacitors in series connected to a power source.
C1=air phase C2=electrolyte phase CT=total system capacitance L = length of phase κ= dielectric constant Vo = battery voltage C=[itex]\frac{κ*ε*A}{d}[/itex] C1=[itex]\frac{ε*A}{L}[/itex] C2=[itex]\frac{κ*ε*A}{L}[/itex] CT=[itex]\frac{C1*C2}{C1+C2}[/itex]=[itex]\frac{κ*ε*A}{(1+κ)*L}[/itex] Q=CT*Vo=[itex]\frac{κ*ε*A*Vo}{(1+κ)*L}[/itex] Q is conserved in series capacitors so V1=Q/C1=[itex]\frac{κ*Vo}{(1+κ)}[/itex] V2=Q/C2=[itex]\frac{Vo}{(1+κ)}[/itex] The voltage drop across the electrolyte phase should therefore be from V2 to zero Based on this, I rederived the potential function using Laplace's Eqn I1=Bx+C I2=Dx+P Considering the system from 0<x<2L where x=0 corresponds to ground and x=2L is the positive terminal facing the air phase... I1=Bx2BL+Vo I2=Dx Both these functions must equal the potential at their interface, e.g. V2 found above. After a bit of algebra you end up with I1=[itex]\frac{κ*Vo}{(1+κ)*L}[/itex]*(x2L)+Vo I2=[itex]\frac{Vo}{(1+κ)*L}[/itex]*x But I'm uncomfortable with this solution as it violates the BC stipulated by conservation of charge that I describe in the original post. E2 should be equal to zero, but this is clearly not the case if we take the derivative of I2... [itex]\nabla[/itex]I2=[itex]\frac{Vo}{(1+κ)*L}[/itex]=/=0 So... does anyone know where this discrepancy arises from? Is it fundamentally flawed to model this as a series of capacitors? Or does the E found in this way reflect the E in the double layer??? Any help would be greatly appreciated. 


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