A capacitor consisting of two parallel plates of area A = 11.30 m2 and spacing d = 0.19 m is filled with two parallel slabs of dielectric material of equal thickness and with dielectric constants κ1 = 2.10 and κ2 = 8.50, respectively. What is the total capacitance?(parallel, not side by side)
2007-03-08 12:59:56 · 3 answers · asked by 2 2 1 in Science & Mathematics ➔ Physics
You would treat this as if there were two capacitors in series.
the capacitor of the first is C1=Aεκ1/(d/2) and of the second
C2=Aεκ2/(d/2). C1=11.3*8.85E-12*2.1/0.095=2.211E-9
C2=11.3*8.85E-12*8.5/0.095=8.948E-9
Ctotal=C1C2/(C1+C2)=1.73 nF
2007-03-08 13:37:28 · answer #1 · answered by Rob M 4 · 2⤊ 0⤋
For your future need for this type of question, that formula can be found in the ARRL Amateur radio Handbook. By the way, parallel, is side by side, think about it.
2007-03-09 05:31:42 · answer #2 · answered by Anonymous · 0⤊ 0⤋
C(k1) = ((k1)*(A)/(d/2)) = ((2.1)*(11.3)/(0.095)) = 2.2nF
C(k2) = ((k2)*(A)/(d/2)) = ((8.5)*(11.3)/(0.095)) = 9.0nF
Ctotal = (C(k1))*(C(k2))/(C(k1))+(C(k2))
= (2.2nF)*(9.0nF)/(2.2nF)+(9.0nF)
= 1.77nF
2007-03-12 11:38:46 · answer #3 · answered by joshnya68 4 · 1⤊ 0⤋