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Parallel plate capacitor and dielectric

  1. Jun 9, 2008 #1
    1. The problem statement, all variables and given/known data
    a parallel-plate capacitor (2 plates) is filled with titanium dioxide which has a dielectric constant of 80. the area of each plate is 1 cm^2.

    a) what must be the spacing between the plates if one wishes to make a 10^-10 farad capacitor?

    b) if the dielectric strength of titanium dioxide is 10^8, what is the maximum chrage that this capacitor can safely store?
    2. Relevant equations

    dielectric constant kappa = 80
    epsilon_0 -- constant = 8.85*10^-12

    capacitance C = kappa[(epsilon_0*A)/d] where A is area, given = 1 cm sq, d is distance, unknown.

    3. The attempt at a solution

    part a:

    capacitance C = kappa[(epsilon_0*A)/d] where C = 10^-10 farad, kappa = 80, epsilon_0 = 8.85*10^-12 ---> solve for d, d = 0.0708 meters = 7.08 cm

    correct?

    part b:

    unsure, how does dielectric strength relate to dielectric constant?
    what equation should i use, is it similar to one the i used in part a?

    thanks
     
  2. jcsd
  3. Jun 9, 2008 #2

    rl.bhat

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    Dielectric strength is the maximum electric field that an be applied across the plates before the dielectric breaks down.
     
  4. Jun 9, 2008 #3

    alphysicist

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    Scholio,

    Part a does not look correct to me. (Did you convert the area correctly to m^2? It is not 0.01 m^2.)
     
  5. Jun 9, 2008 #4
    alphysicist, isn't 0.01m = 1cm? do i need to square it?

    also i tried part b, assuming dielectric strength(= 10^8 V/m) = max electric field, i used this eq:

    electric field E = kq/r^2 where k is constant 9*10^9, r is distance -- from part a =0.0708m, and then solved for q

    so q = E*r^2/k = [(10^8)(0.0708^2)]/(9*10^9) = 5.57*10^-5 coulombs

    correct?
     
  6. Jun 9, 2008 #5

    alphysicist

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    That's right. 1 cm = 0.01m, but what you have here is 1 cm^2, which you might think of as 1 cm cm. A couple of different ways to think the conversion:

    [tex]
    1 cm^2 \left(\frac{1 m}{100 cm}\right)^2 = 1\times 10^{-4} m^2
    [/tex]

    or equivalently:

    [tex]
    1 cm\cdot cm \left(\frac{1 m}{100 cm}\right) \left(\frac{1 m}{100 cm}\right) = 1\times 10^{-4} m^2
    [/tex]



    A capacitor has its charge spread out over a surface. The charge is not a point charge, and so you cannot use Coulomb's law kq/r^2. For a parallel plate capacitor, we say that the electric field is (approximately) constant. What's the relationship between potential and electric field when the field is constant?
     
  7. Jun 9, 2008 #6
    i changed my area from 0.01 to 1*10^-4m^2 and got a distance, for part a, to be 7.08*10^-4 meters

    part b:
    as for be, i want to use the eq: capacitance C = epsilon_0*A/d where epsilon_0 = 8.85*10^-12, A = 1*10^-4 m^2, d = 7.08*10^-4m

    but the equation is pretty useless if i know all the unknowns. i need an equation with electric field E = dielectric strength because that is given as 10^8 V/m

    is potential = electric field when the field is constant? i'm not too sure, couldn't find an equation relating the two.
     
  8. Jun 9, 2008 #7

    alphysicist

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    I noticed that you had the relationship between field and potential in another thread:

    [tex]
    \Delta V = - \int\limits_a^b \vec E \cdot d\vec r
    [/tex]

    What does the right side simplify to when E is a constant field?

    Once you have that, how is V and Q related for a capacitor?
     
  9. Jun 9, 2008 #8
    if i did the integral holding E constant would i get r_b - r_a for the right side?

    as for the relationship with V and q for a capacitor, capaciance C = Q/V so if i sub in r_b - r_a for V in the C eq, i'd get C = Q/ (r_b - r_a)

    correct?
     
  10. Jun 9, 2008 #9

    alphysicist

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    To be more precise, the E would come out of the integral, and r_b-r_a is the length being integrated over, so the magnitude of the potential would be:

    [tex]
    \Delta V = E d
    [/tex]

    (The general formula for a constant electric field is:

    [tex]
    \Delta V = - E d \cos(\theta)
    [/tex]

    where [itex]\theta[/itex] is the angle between the field and displacement.)

    So you know here that V= Ed, which as you say is used in C=Q/V.


     
  11. Jun 9, 2008 #10
    so since V = Ed, and C = Q/V ---> C = Q/Ed where C = 10^-10 farad, E = 10^8 V/m, d = 7.08*10^-4 meters

    so Q = 7.08*10^-6 coulombs, seems a little small

    correct?
     
  12. Jun 9, 2008 #11

    alphysicist

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    That looks right to me.
     
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