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phyzmatix
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[SOLVED] Parallel-plate capacitor: Two dielectric materials
A parallel-plate capacitor with area A = 5.56 cm^2 and separation d = 5.56 mm has the left half of the gap filled with material of dielectric constant K1 = 7.00 and the right half filled with material of dielectric constant K2 = 12.0. What is the capacitance?
[tex]C = \frac{\kappa \epsilon_{0} A}{d} [/tex]
[tex] \frac {1}{C_{tot}} = \frac {1}{C_{left}} + \frac {1}{C_{right}}[/tex]
Since it's one capacitor (two plates) and since the dielectric materials are next to each other (as opposed to stacked) I approached the problem as if the two halves of the capacitor were separate parallel capacitors and calculated:
[tex]C_{left} = 3.10 pF [/tex]
[tex]C_{right} = 5.31 pF [/tex]
[tex]C_{tot} = 1.96 pF [/tex]
But this is not amongst the given possible answers. Where did I go wrong?
Cheers
phyz
Homework Statement
A parallel-plate capacitor with area A = 5.56 cm^2 and separation d = 5.56 mm has the left half of the gap filled with material of dielectric constant K1 = 7.00 and the right half filled with material of dielectric constant K2 = 12.0. What is the capacitance?
Homework Equations
[tex]C = \frac{\kappa \epsilon_{0} A}{d} [/tex]
[tex] \frac {1}{C_{tot}} = \frac {1}{C_{left}} + \frac {1}{C_{right}}[/tex]
The Attempt at a Solution
Since it's one capacitor (two plates) and since the dielectric materials are next to each other (as opposed to stacked) I approached the problem as if the two halves of the capacitor were separate parallel capacitors and calculated:
[tex]C_{left} = 3.10 pF [/tex]
[tex]C_{right} = 5.31 pF [/tex]
[tex]C_{tot} = 1.96 pF [/tex]
But this is not amongst the given possible answers. Where did I go wrong?
Cheers
phyz