Capacitance for a capacitor with two dielectrics

  • #1
VelocityOfTheSound
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Homework Statement:
If I have two parallel conductive plates, that is, a capacitor, with two dielectrics k1 and k2 between the plates, and I want to know how much is the capacitance, knowing that I can solve the problem finding the equivalent capacitance for the two capacitors, one with k1 and the other with k2, how to determine whether they are in series or in parallel?
Relevant Equations:
kQ = CV
The geometry of the capacitor can be either cylindrical or spherical.
 

Answers and Replies

  • #2
bob012345
Gold Member
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If you already know the what the total capacitance looks like then you can test either alternative and see which one matches of course. If you don't you can think of what physical property must be in series and go from there.
 
  • #3
Delta2
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It depends on the exact problem setup (for which you are a bit vague I must admit).

If we have a parallel plate capacitor with ##d## the distance between the plates and ##l## the length of the plates (and ##w## the depth of the plates) and has two dielectrics between the parallel plates then(assuming the capacitor plates are up and down):
  • if one dielectric is a slab with dimensions ##\frac{d}{2} \times l\times w## and the other also a slab of the same dimensions this means that one dielectric is in top of the other and then you have two capacitors in series. The two capacitors have area ##A=l\times w##, distance between plates ##\frac{d}{2}## and one is with dielectric ##k_1## and the other with dielectric ##k_2##
  • if one dielectric is a slab with dimensions ##d\times \frac{l}{2} \times w## and the other again the same and each dielectric slab is next to the other. The two capacitors are in parallel now, each capacitor has now area ##A=\frac{l}{2}\times w##, but distance between the plates ##d## and one is with dielectric ##k_1## and the other with dielectric ##k_2##.
  • the case that each dielectric slab is ##d\times l\times \frac{w}{2}## is similar to the second case.
 

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