How Much Charge Can a Capacitor Hold Before Failure?

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SUMMARY

The discussion focuses on calculating the charge capacity of a parallel plate capacitor using specific parameters: side length (L), dielectric thickness (d), and dielectric strength (Eb). The relevant equations include C = kEo(A/d) for capacitance and q = CV for charge. The user calculated a capacitance of 3.096 x 10^-9 F and a charge capacity of 4.96 x 10^-4 C using a dielectric constant (k) of 3.5 and dielectric strength (Eb) of 16. However, there was confusion regarding the assumption that electric field strength (E) equals dielectric strength (Eb) in the charge calculation.

PREREQUISITES
  • Understanding of parallel plate capacitor theory
  • Familiarity with capacitance equations, specifically C = kEo(A/d)
  • Knowledge of dielectric materials and their properties
  • Basic algebra and unit conversions in physics
NEXT STEPS
  • Research the implications of dielectric strength on capacitor performance
  • Learn about the effects of different dielectric materials on capacitance
  • Explore advanced capacitor design techniques for increased charge capacity
  • Study the relationship between electric field strength and dielectric breakdown
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Students in electrical engineering, physics enthusiasts, and professionals designing capacitors or working with dielectric materials will benefit from this discussion.

vysero
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Homework Statement



You are building a parallel plate capacitor out of two square metal plates with side lengths of L and a thin piece of dielectric of thickness d and dielectric strength of Eb. The piece of dielectric will act as a means of separating the two metal plates. In terms of l, Eo(epsilon knot), k, d, Eb how much charge can the capacitor hold before it fails? With l = 10cm, d = .1mm and paper as a dielectric what is the capacitance of your capacitor? How much charge can the capacitor hold before it fails?[/B]

Homework Equations



eq1: C = EoL; L for a parallel-plate capacitor is = A/d

with a dielectric completely filling the space between the plates:[/B]

eq 2: C = kEo(A/d)
eq 3: q = CV
eq 4: V = E/d


The Attempt at a Solution



Using those equations I came up with:

q = CV = (kEoL^2/d)(Eb/d) = EbkL^2Eo/d^2

For the second part of the question with k = 3.5 and eq2 I come up with:

C = 3.5(8.85x10^(-12))(.1^2/.0001) = 3.096 x 10^-9 F

and for the final equation where Eb = 16 I said:

q = 16(3.5)(.1)^2(8.85x10^(-12))/(.0001)^2
= 4.96x10^(-4) C

However, I made the assumption that E = Eb in eq3. If someone could check my work and tell me if I did this problem correctly that would be great, any help appreciated.[/B]
 
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vysero said:

Homework Statement



You are building a parallel plate capacitor out of two square metal plates with side lengths of L and a thin piece of dielectric of thickness d and dielectric strength of Eb. The piece of dielectric will act as a means of separating the two metal plates. In terms of l, Eo(epsilon knot), k, d, Eb how much charge can the capacitor hold before it fails? With l = 10cm, d = .1mm and paper as a dielectric what is the capacitance of your capacitor? How much charge can the capacitor hold before it fails?[/B]

Homework Equations



eq1: C = EoL; L for a parallel-plate capacitor is = A/d

with a dielectric completely filling the space between the plates:[/B]

eq 2: C = kEo(A/d)
eq 3: q = CV
eq 4: V = E/d


I do not understand what you mean with eq 1, and the last equation (eq 4) is wrong . Check.
 

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