#1
Jun1408, 05:00 AM

P: n/a

If two electrodes are sandwitching two dielectric materials with very
different dielectric constants (but the same thickness), say, water and glass. Would the new dielectric constant lies in between the original two? What would be the electric field in between the dielectric materials? I suppose not half of the total electric field imposed by the electrodes. Would the larger dielectric constant material take up more of it? If the water contains ions, would that change its dielectric constant from that of its pure form (about 80)? Is there any relation between dielectric constant and dielectric strength? 


#2
Jun1508, 05:00 AM

P: n/a

rambotrout wrote:
> > If two electrodes are sandwitching two dielectric materials with very > different dielectric constants (but the same thickness), say, water > and glass. Would the new dielectric constant lies in between the > original two? > > What would be the electric field in between the dielectric materials? > I suppose not half of the total electric field imposed by the > electrodes. Would the larger dielectric constant material take up more > of it? > > If the water contains ions, would that change its dielectric constant > from that of its pure form (about 80)? Water with ions is electrically conductive. A better example would be two solid slabs, perhaps contrasting polyethylene foam (about 1.3, coax cable) and poly(vinylidene fluoride) at 12.2 or potassium tantalate niobate at 6000. What if you insulated your DC electrodes with a couple of microns thickness of ParyleneC film then dipped them in electrolyte solution or placed a copper slab inbetween? > Is there any relation between dielectric constant electric field attenuation > and dielectric > strength? breakthrough voltage/thickness  Uncle Al http://www.mazepath.com/uncleal/ (Toxic URL! Unsafe for children and most mammals) http://www.mazepath.com/uncleal/lajos.htm#a2 


#3
Jun1608, 05:00 AM

P: n/a

Please correct me if I am wrong.
Say the voltage across the slabs is Vt, then Vt = V1 + V2, where V1, V2 are voltages across the two different dielectric materials. Therefore, Vt = Q*d1/(E1 * A) + Q*d2/(E2 * A), where Q = charge in Coulomb, d = thickness of material, E = dielectric constant, and A = area. 


#4
Jul2208, 05:00 AM

P: n/a

voltage across the slabs is Vt
On Sun, 15 Jun 2008 19:01:13 +0000, rambotrout wrote:
> Please correct me if I am wrong. > > Say the voltage across the slabs is Vt, then Vt = V1 + V2, where V1, V2 > are voltages across the two different dielectric materials. Therefore, > > Vt = Q*d1/(E1 * A) + Q*d2/(E2 * A), > > where Q = charge in Coulomb, d = thickness of material, E = dielectric > constant, and A = area. You are on the right track. In order to calculate the equivalent dielectric constant for the whole assembly, you can treat the system as two capacitors in series, each with a single kind of dielectric. I usually imagine an infinitesimally think conductor between them. Remember that capacitors in series add in reciprocals (1/C_eq = 1/C_1 + 1/ C_2), and you can work out what the effective dielectric constant is (which would be relatively simple if d1 == d2, by the way). 


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