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Capacitance for a capacitor with two dielectrics
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[QUOTE="Delta2, post: 6508345, member: 189563"] 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): [LIST] [*]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. [/LIST] [/QUOTE]
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Capacitance for a capacitor with two dielectrics
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