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## Homework Statement

A parallel plate capacitor has area A = 1 cm

^{2}and a plate separation of d = 0.01m (1cm). Water at room temp (20°C) is poured into a mica cylinder and placed between the plates filling the volume of 1cm3. Find the Maximum capacitance, voltage and charge for the capacitor as well as which dielectric will break down first.

## Homework Equations

Q=CV

V = Ed

mica

cyclinder thickness = 0.25cm x 2 (touches both sides of the capacitor =0.5cm = 0.005m)

dielectric constant = 3-6 (3)

dielectric breakdown = E = 118MV/m

water

thickness = d = 0.005m

electric constant =

*ε*0 = 8.854×10−12 F

dielectric constant = K = 80.1

dielectric breakdown = 65MV/m

Capacitance for 2 capacitors in series:

$$C_T\frac{C_1\times C_2}{C_1+C_2}$$

Breakdown voltage the capacitor:

$$V{max}=\frac{V_{bd}}{\frac{C_T}{C_1}}$$

## The Attempt at a Solution

In a two dielectric capacitor, if one dielectric breaks down the capacitor fails. To solve for the voltage across each capacitor in series, first find the capacitance:

https://www.electronics-tutorials.ws/capacitor/cap_7.html

**Series capacitor specs:**

$$C=K\varepsilon_0\frac{A}{d}$$

Max voltage for mica:

$$(118\times10^6V)\times0.005m=590\times10^3V=590kV$$

Max voltage for water:

$$(65\times10^6V)\times0.005m=325\times10^3V= 325kV$$

Capacitance for mica:

$$C=K\varepsilon_0\frac{A}{d}=C=\frac{(8.85\times10^{-12})\times3\times0.0001}{0.005}=5.31\times10^{-13}F=0.531pF$$

Capacitance for water:

$$\frac{(8.85\times10^{-12})\times80.1\times0.0001}{0.005}=1.41777\times10^{-11}F=14.1777pF=14.2pF$$

Charge for mica:

$$(118\times10^6V)\times(5.31\times10^{-13}F) =6.2658\times10^{-5}C=626.58nC=627nC$$

Charge for water:

$$(65\times10^6V)\times1.41777\times10^{-11}F=5.529303\times10^{-8}=552.9303nC=553nC$$

(this equation can be used when only two capacitors are in series)

$$C_T\frac{C_1\times C_2}{C_1+C_2}$$

$$C_T=\frac{(1.41777\times10^{-11}F)\times (5.31\times10^{-13}F)}{(1.41777\times10^{-11}F)+(5.31\times10^{-13}F)}=\frac{(7.528359\times10^-24) F}{(1.47087\times10^-11)F}=5.11830\times10^{-13} = 0.511830pF=0.512pF$$

$$V_{C_{1}}=\frac{C_T}{C_1}\times V_T$$

I modified this equation to solve for the max voltage across the mica's breakdown voltage(bd1):

$$V_{bd1}=\frac{C_T}{C_1}\times V{max}$$

$$V{max}=\frac{V_{bd}}{\frac{C_T}{C_1}}$$

mica:

$$V_{c1}max=\frac{59\times10^4V}{\frac{0.511830pF

}{0.531pF}}=\frac{59\times10^4V}{0.963898pF}=612,097.7668371V=612.1kV$$

water:

$$V_{c2}max=\frac{325\times10^3V}{\frac{0.511830pF

}{14.1777pF}}=\frac{325\times10^3V}{0.036101pF}=9002505.71V=9MV$$

The mica containing the water will break down first.