1) A parallel-plate capacitor, of area 5.70×10-2 m2, has the gap between the pla

Question

1) A parallel-plate capacitor, of area 5.70×10-2 m2, has the gap between the plates filled by benzene, a dielectric. If a laboratory meter measures the capacitance to be 440 pF, what is the capacitance when there is only air in the gap?

2) What is the separation of the plates?

3)The dielectric between the plates is now replaced by two media, in different proportions. If the capacitor is viewed edge-on as shown, Teflon fills the lower volume between the plates, to a fraction of 0.45 of the total, and Mylar fills the remainder. What does the meter read now? (Note: This capacitor can be treated as two capacitors in parallel; the total capacitance is the sum of the capacitances of the two parts, each evaluated for its own geometry.)

Explanation / Answer

relation betweeen capacitence and dielectric constant is

C' = CK

C= C'/K = 440pF/2.3 = 191.3 pF

formula for ccapacitence of the parallel plate capacitor is

C= eo A/d

d = eo A/ C

= 8.85 * 10^-12 ( 5.70 * 10^-2 )/191.3* 10^-12

=2.63 * 10^-3 m

V= A d = 5.7 * 10^-2 ( 2.63 * 10^-3) = 14.991 * 10^-5 m^3

Are of teflon = 5.7 * 10^-2 ( 0.45) = 2.565 * 10^-2 m^2

Area of mylar = 5.7 * 10^-2 ( 0.55) = 3.135* 10^-2 m^2

C net =k_t eo A_t / d+ k_M eo A_M/d

= 8.85 * 10^-12 /2.63 * 10^-3 m( 2.1 ( 2.565 * 10^-2 m^2) + 3.1 (  3.135* 10^-2 m^2)

=0.508 nF

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