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## Homework Statement

there's a paper insulated plate capacitor.

a.) What should the area of the plates be when the capacitance is 0,1 μF.

thickness of the paper in between the plates of the plate capacitor is 0,05mm

the relative permittivity of the paper is 2,5

b.) How big of a charge can be brought to this capacitor, if the dielectric strength of the paper is 4*10^7 Volts/meter

## Homework Equations

i [/B]##E_{max} * d = U_{max}##, where E= dielectric strength, U= breakdown voltage ( I got these English terms from wikipedia article by switching the language settings, so hopefuly the formula is sensible)

##C=Q/U##

##ε_0= 8,85*10^{-12} \frac{F}{m}##

##C= \frac{ε_{r}ε_{0}A}{d}##

## The Attempt at a Solution

In the A part, it was assumed I think that the distance of between plates of the capacitors was "fixed" at that value. Also the entire space between plates was filled with the insulator paper I reckon.

A.) this part is straightforward because the formulas and known values are enough to work with

##A=\frac{Cd}{ε_{r}ε_{0}}##

##A≈0,226m^2##I was looking at my notes about deriving the Capacitance formula from earlier work. But I remained a little bit confused about the Area term of the formula.##E= \frac{σ}{ε_0ε_r} = \frac{ \frac{Q}{A}}{ε_0ε_r}= \frac{Q}{Aε_0ε_r}##

##U=Ed##

##U=\frac{Qd}{ε_0ε_rA}##

##C=Q/U##

##C= \frac{Q}{\frac{Qd}{ε_0ε_rA}}##

##C= \frac{ε_{r}ε_{0}A}{d}##

question what I don't really remember is that is this resulting area like...the combined area of the both plates, negatively charged plate area plus positively charged area? I think it's just one of the plate's area, because charge density is negative value for the negative plate. And charge density is positive value for positive plate. It wouldn't really make sense other ways.

B.)

It is known that the dielectric strength was 4*10^7 V/m

also I think we are supposed to utilize the paper's thickness as the value for d (but I'm not 100% certain about it... this seemed to be the way to go?)

##E_{max} *d= U_{max}##

## U_{max}= 4*10^7 V/m*5*10^{-5}m

U_max= 2000Volts##

Can you simply calculate that

##C= Q/U_{max}##

##C*U_{max}= Q##

## 10^{-7} C/V * 2000V##

##Q=2*10^{-4} Coulombs##

My confusion at the B part was about the Capacitance... Are you supposed to calculate with the capacitance value of ##10^{-7}## farads? Why is the capacitance the same value?(the same capacitance as waas in the earlier A part)

In the A part, it was assumed I think that the distance of between plates of the capacitors was "fixed" at that value. Also the entire space between plates was filled with the insulator paper I reckon.

I guess the idea about the capacitance being equal makes sense conceptually if I look at the formula for

##C= \frac{ε_{r}ε_{0}A}{d}##

because in both parts a) and b) of the assignment the Epsilon_0 and Epsilon_r are the same values betweeen each part. And I think also the d=distance is the same value for a) and b) parts of the assignment. Same with the area also.

But I don't really see how the other equation (C=Q/U) ties into that thing above. I suppose I'm conceptually confused at this point.