# Parallel Plate Capacitor Conundrum

 P: 4 1. The problem statement, all variables and given/known data Hi everyone, I'm new here at posting but have found this forum very useful in the past. I have hit a snag with a capacitor problem... any thoughts? Assume you have a parallel plate capacitor separated by one cm of air and one cm of a dielectric with a dielectric constant of six. What fraction of the voltage between the two plates is across the air? Look up the breakdown strength of air and calculate the voltage when the air will begin to discharge. ε = 6 ε0= 1.0059 d = 1 cm Breakdown strength of air is 3MV/m 2. Relevant equations V= I X X = 1/jωC = d/(j ε ε0A) V= I d / j ω A Ɛ Ɛ_0 3. The attempt at a solution I started trying to add the capacitances of the two dielectics together in series, however, can I assume that the capacitance will divide according to their dielectric constants? That is the voltage gradient will be 6 times stronger over the air compared to the dielectric?
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P: 11,865
 Quote by James11111 1. The problem statement, all variables and given/known data Hi everyone, I'm new here at posting but have found this forum very useful in the past. I have hit a snag with a capacitor problem... any thoughts? Assume you have a parallel plate capacitor separated by one cm of air and one cm of a dielectric with a dielectric constant of six. What fraction of the voltage between the two plates is across the air? Look up the breakdown strength of air and calculate the voltage when the air will begin to discharge. ε = 6 ε0= 1.0059 d = 1 cm Breakdown strength of air is 3MV/m 2. Relevant equations V= I X X = 1/jωC = d/(j ε ε0A) V= I d / j ω A Ɛ Ɛ_0 3. The attempt at a solution I started trying to add the capacitances of the two dielectics together in series, however, can I assume that the capacitance will divide according to their dielectric constants? That is the voltage gradient will be 6 times stronger over the air compared to the dielectric?
Hello James11111, Welcome to Physics Forums.

What's the relationship between two capacitors of identical dimensions when one has no dielectric (air's dielectric constant is negligibly different from that of vacuum) and one that is filled with a dielectric? How does voltage divide across series capacitors?
 P: 4 Yep figured the first bit. It becomes a simple voltage divider circuit. k1=1.0006 k2=6.0 Vout = [Z2/(Z1+Z2)]*Vin Zx = 1/j$\omega$Cx - Sub in for each Z Cx = kx * ε0*A/d - Sub in for each C Vout = [k1/(k1+k2)]*Vin - After cancelling and simplifying. The fraction of voltage over the dielectric component after substituting in k values is: Vout = [1/(1+6)]*Vin=1/7*Vin Which means that the fraction of voltage over the air must be 6/7.
 P: 4 Parallel Plate Capacitor Conundrum Now to work out at what voltage the air will breakdown and discharge. The breakdown strength of air is 3 MV/m, right? This maybe too simplified but can I do the following? distance between plates = 0.02 m breakdown voltage of air over 0.02 m is 3M * 0.02 = 60,000 V the fraction of 60,000V stressed across the air gap of the capacitor 6/7. therefore, the air will begin to discharge when the voltage reaches: VDischarge = (6/7)*60,000 = 51,428.6V
Mentor
P: 11,865
 Quote by James11111 Now to work out at what voltage the air will breakdown and discharge. The breakdown strength of air is 3 MV/m, right? This maybe too simplified but can I do the following? distance between plates = 0.02 m breakdown voltage of air over 0.02 m is 3M * 0.02 = 60,000 V
Whoops. The air gap is not 2 cm wide -- half of that space is filled with dielectric. The 'air capacitor' gap is half that width. The dielectric surface serves as one plate of the air capacitor.
 the fraction of 60,000V stressed across the air gap of the capacitor 6/7. therefore, the air will begin to discharge when the voltage reaches: VDischarge = (6/7)*60,000 = 51,428.6V
 P: 4 Aha, Cheers!

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