Question

two 20* 20 cm capacitor plates are separated by 2 mm and connected to a 100v dc source.A 1-mm-thick sheet of dielectric is inserted between the plates giving a 1/2-s pulse of 30nA average current.What is the permittivity of the dielectrics sheet?

Answer 1

I think the first step will be to determine the capacity of the parallel-plate capacitor without the dielectric sheet. You have all the information to do that. Once you have the capacity, what's the total charge on the plates?

Once the dielectric sheet is between the plates, the capacitance will be changed. The new capacitance can be determined by first calculating the new charge on the plates (remember: a current means a change in charge).

Again using the formula for the capacitance of a parallel plate capacitor, you can find the permittivity of sheet plus a leftover air-gap (!).

Hope this will get you a bit further.

Answer 2

ok i did it this way \[30nA \times 0.5\sec=15nC\],\[C =15nC \div 100V=150pF\],\[C=C_{1}C_{2}/(C_{1}+C_{2})\]=\[\epsilon_{1} \epsilon A/d \times \epsilon_{2} \epsilon A/d/(\epsilon_{1} \epsilon A/d + \epsilon_{2} \epsilon A/d)\]=\[\epsilon_{1} \epsilon_{2}\epsilon A/((\epsilon_{1}+ \epsilon_{2})d)\],\[150pF=1\times \epsilon_{2}\times8.85p \times0.04/((1+\epsilon_{2})\times0.001)\],which eventually gives \[\epsilon\]=0.7352 which is not appropriate and the answer say it is 12

Answer 3

Assume d = 2 mm, A=( 2 cm * 2 cm ), and epsilon_0 = 8.854 pF/m
Without the dielectric slab in the capacitor, the total CHARGE on the capacitor plates would be:
\[Q_0=CV=\epsilon_0A/d * V \]
When the dielectric is inserted, a current flows and adds a charge to the capacitor plates, so the new charge of the capacitor will be:
\[Q_1=Q_0 + Q_{add} = \epsilon_0AV/d + I_{av}\Delta{}t \]
Using this charge, you can determine the new capacitance and insert that capacitance into your equation for two capacitors in series (as C).

Using this approach, you should get and epsilon_r of 12.

Answer 4

I worked out \[\epsilon_{r}\] = 12.111 Assume the totall C = 150pF is a mistake the correct equaltion is \[(\epsilon_{1} \epsilon_{2} \epsilon A-\epsilon_{1} ^{2}\epsilon A)/(2(\epsilon_{1} +\epsilon_{2})d) =150pF\].

Answer 5

thanks for the hint