4. [1pt] A parallel-plate capacitor, of area 6.16×10 -2 m 2 , has the gap betwee

Question

4. [1pt]
A parallel-plate capacitor, of area 6.16×10-2 m2, has the gap
between the plates filled by Teflon, a dielectric. If a
laboratory meter measures the capacitance to be 510 pF, what
is the capacitance when there is only air in the gap?

Correct, computer gets: 2.43e+02 pF

5. [1pt]
What is the separation of the plates?

6. [1pt]
The dielectric between the plates is now replaced by two media, in different proportions. If the capacitor is viewed edge-on as shown, Mylar fills the lower volume between the plates, to a fraction of 0.20 of the total, and glycerin 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.)

Answer: Last Answer: 1.07e-8 F
Incorrect, tries 4/10.

Dielectric Constant at 20 °C Teflon 2.1 polyethylene 2.25 benzene 2.28 Mylar 3.1 Plexiglas 3.4 neoprene 6.7 glycerin 42.5 water 80.4

Explanation / Answer

4) C = 510 pF , A =6.16*10^-2 m^2

k = 2.1

C= kCo

Co = C/k =510/2.1= 2.43*10^2 pF

5) Co =Aeo/d

d = Aeo/Co =(6.16*10^-2*8.85*10^-12)/(2.43*10^-10)

d = 2.25 mm

6) C1 = Ak1eo/0.2d

C2 = Ak2eo/0.8d

C =C1+C2

C = (Aeo/d)[k1/0.2+k2/0.8]

C = (2.43*10^-10) ((3.1/0.2)+(42.5/0.8))

C =1.67*10^-8 F

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