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## Homework Statement

A parallel plate capacitor has area A = 1 cm

^{2}and a plate separation of d = 0.01m. Water at room temp (20°C) is poured into a ziplock bag and placed between the plates filling the volume of 1cm

^{3}. Find the maximum capacitance, voltage and charge for the capacitor.

## Homework Equations

Q=CV

ziplock:

bag thickness = 1.5mm x 2 (touches both sides of the capacitor = 3mm = 0.003m

dielectric constant = 2.3

dielectric breakdown = 25MV/m

water:

thickness = 0.007m

dielectric constant = 80.1

dielectric breakdown = 65MV/m

Capacitance for two dielectrics in series:

$$C=\frac{\varepsilon_0A}{\frac{d_1}{k_1}\frac{d_2}{k_2}}$$

Voltage for two dielectrics in series:

$${V_r} = {E_{r1}}{d_1} + {E_{r2}}{d_2}$$

## The Attempt at a Solution

Maximum capacitance:

$$\frac{(8.85\times10^{-12})\times0.0001}{\frac{0.003}{2.3}+\frac{0.007}{80.1}}=6.35895\times10^{-13}F=0.635895nF=$$Maximum voltage:

$$(65\times10^6V\times0.007m)+(25\times10^6V\times0.003m) =(45.5\times10^3V)+(75\times10^3V) = 455kV+75kV=530kV$$Maximum charge:

$$(6.35895\times10^{-13}F)\times(53\times10^{4}V)=(3.370243\times10^{-7}C)=337.0243nC$$

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