Dielectric strength and breakdown voltage capacitor

In summary: C?##In summary, it can be assumed that the electric field is equal to the dielectric strength of the material the capacitor is made of. If the dielectric strength is 4*10^7 V/m, then the electric field is 2000 V. The capacitor can hold a charge of 20,001*10^{-5} C.
  • #1
late347
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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.
 
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  • #2
late347 said:
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?

The area of one plate.
 
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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.
 
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  • #4
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TJGilb said:
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.

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
TJGilb said:
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.
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.
 

1. What is dielectric strength and breakdown voltage of a capacitor?

Dielectric strength refers to the maximum electric field that a material can withstand without breaking down. Breakdown voltage, on the other hand, is the minimum voltage required for a dielectric material to break down and allow current to flow through. In a capacitor, these two properties determine its ability to store and withstand electric charge.

2. How is the dielectric strength of a capacitor measured?

The dielectric strength of a capacitor is typically measured by applying a gradually increasing voltage until the material breaks down and allows current to flow. This voltage is then recorded as the breakdown voltage for that particular material and capacitor design.

3. What factors affect the dielectric strength and breakdown voltage of a capacitor?

The dielectric strength and breakdown voltage of a capacitor can be affected by several factors, including the type and thickness of the dielectric material used, the distance between the capacitor plates, and the overall design and construction of the capacitor. Temperature and humidity can also play a role in these properties.

4. Why is the dielectric strength and breakdown voltage important in capacitors?

The dielectric strength and breakdown voltage are important properties in capacitors because they determine the maximum voltage that the capacitor can handle before breaking down and becoming damaged. These properties also impact the overall performance and reliability of the capacitor in various applications.

5. How can the dielectric strength and breakdown voltage of a capacitor be improved?

The dielectric strength and breakdown voltage of a capacitor can be improved by using high-quality dielectric materials with higher dielectric constants and lower porosity. Increasing the distance between the capacitor plates and using proper insulation techniques can also enhance these properties. Additionally, proper design and testing processes can help ensure that the capacitor can withstand the desired voltage without breaking down.

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