# Homework Help: Dielectric strength and breakdown voltage capacitor

1. Dec 10, 2016

### late347

1. The problem statement, all variables and given/known data
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

2. Relevant equationsi
$E_{max} * d = U_{max}$, 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)
$C=Q/U$
$ε_0= 8,85*10^{-12} \frac{F}{m}$
$C= \frac{ε_{r}ε_{0}A}{d}$
3. 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 wouldnt 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.

2. Dec 10, 2016

### TJGilb

The area of one plate.

3. Dec 10, 2016

### TJGilb

So for part b, you have your max potential of 2000 V from the max electric field (dielectric strength). Now you could approach this two ways. On the one hand, you could directly calculate what charge would give you that electric field, or you could assume the same capacitance and calculate from that. It's hard to interpret what assumptions the problem is asking you to make from the problem statement presented.

4. Dec 11, 2016

### late347

666565
ok, if we try to utilize the assumption that the max electric field = dielectric strength = 4*10^7 V/m

According to my notes from my notebook...

In an insulatro filled plate capacitor the following should hold for the electric field strength.
Plate area should stay the same area regardless, I reckon...
$E= \frac{ \frac{Q}{A}}{ε_0ε_r}$
$Q= {EAε_0ε_r}$

$Q= 0,226m^2 ~*~ 4*10^7 \frac{V}{m}~*~8,85*10^{-12}\frac{F}{m} ~*~2,5$

meters cancel out --> you have left FV --> C/V* V--> volts cancel out leaving only coulombs

$Q≈20,001*10^{-5}C$

$Q≈2*10^{-4}C$

same answer as could be achieved by assuming that U_max= 2000 V and that C= basically unchanged value of the 10^{-7} farads,
because E_max was 40MV/m and d= 5*10^{-5}m --> U_max= 2000V

$Q/U_{max} = C$
$Q/U_{max} = \frac{ε_0ε_rA}{d}$
$Q= \frac{ε_0ε_rAU_{max}}{d}$

5. Dec 11, 2016

### late347

ok correct if I'm wrong here
for plate capacitor with insulator filling in the space between plates (fully covered by material with ε_r I think)
$E= \frac{ \frac{Q}{A}}{ε_0ε_r}$
$Q= {EAε_0ε_r}$

you can see that the variables: A and ε_0 and ε_r are actually "fixed variables"

So, that, if you want to maximize the size of Q... you have to maximize the size of E variable (electric field strength)
So, that if you plug into that formula the dielectric strength. That should give you the max charge.