- #1
bigevil
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Homework Statement
Find the capacitance of the parallel plate capacitor with 2 dielectrics below. Given that the parallel plate capacitor has area A = WL and the separation between plates is d.
The Attempt at a Solution
My method is to use integration. First solve for two small capacitors dC1, dC2 and then sum them up in parallel.
[tex] y = \frac{d}{L}x [/tex]
[tex] dC_1 = \frac{\kappa_1 \epsilon_0 WL dx}{d(L-x)} [/tex]
[tex]dC_2 = \frac{\kappa_2 \epsilon_0 WL}{d} \frac{1}{x} dx [/tex]
[tex] dC = \frac{dC_1 dC_2}{dC_1 + dC_2} = \frac{\kappa_1 \kappa_2 \epsilon_0 WL}{d} \cdot \frac{1}{\kappa_2 L + x(\kappa_1 - \kappa_2)} [/tex]
[tex]C = \frac{\kappa_1\kappa_2 \epsilon_0 WL}{d(\kappa_1 - \kappa_2)} \left[ ln (\kappa_2 L + x(\kappa_1 - \kappa_2) \right]_0^L[/tex]
The expression is definitely wrong. If we let the two dielectrics have equal value k, we should expect C to follow that of a single dielectric, C' = kC.