- #1
Destroxia
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Homework Statement
[/B]
The figure shows a parallel-plate capacitor with a plate area A = 5.29 cm^2 and plate separation d = 8.10 mm. The left half of the gap is filled with material of dielectric constant κ1 = 6.50; the right half is filled with material of dielectric constant κ2 = 11.4. What is the capacitance?
http://edugen.wileyplus.com/edugen/courses/crs7165/art/qb/qu/c25/fig25_48_wiley.gif
(Given)
A = 5.29 cm^2
d = 8.10 mm
k1 = 6.50
k2 = 11.4
e0 = 8.85E-12
(unknowns)
CNoDielectrics = ?
CDielectrics = ?
Homework Equations
C[/B]Parallel-Plate = (e0*A)/d
CAddedDielectric = CParallel-Plate*Kdielectricconstant
The Attempt at a Solution
CNo Dielectric/2 = (e0*((5.29E-2)/2))/(8.10E-3) = 2.89E-11
Since the separated capacitor is in series (I think), you can use the law of inverse capcitance sums, after multiplying the CNo Dielectric/2 once by k1, and another time by k2, and add them, then inverse them, you get
1/((CNo Dielectric/2)*k1) + 1/((CNo Dielectric/2)*k2) =
= 3.04E-11 + 5.32E-11 = 8.36E-11
1/8.36E-11 = 1.20E-12 = CDielectrics