[1pt] A parallel-plate capacitor, of area 4.51×10 -2 m 2 , has the gap between t

Question

[1pt]
A parallel-plate capacitor, of area 4.51×10-2 m2, has the gap
between the plates filled by glycerin, a dielectric. If a
laboratory meter measures the capacitance to be 330 pF, what
is the capacitance when there is only air in the gap?

Answer:  

5.  
What is the separation of the plates?

6.  
The dielectric between the plates is now replaced by two media, in different proportions. If the capacitor is viewed edge-on as shown, neoprene fills the lower volume between the plates, to a fraction of 0.60 of the total, and polyethylene 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.)

Use this table: Recall that 1 picofarad = 1 pF = 1×10-12 F.

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
Ly

Explanation / Answer

C = k C0

where C0 = capacitance without dielectric (when only air gap)

C = with dielectric.

C0 = 330/ 42.5 = 7.76 pF ....Ans

5. C0 = e0 A / d

(7.76 x 10^-12) = (8.854 x 10^-12) (4.51 x 10^-2) / d

d = 0.0514 m OR 5.14 cm

6. for upper, C1 = k2 (0.4C0) = 6.984 x 10^-12 F

and for lower, C2 = k1 (0.60 C0) = 31.196 x 10^-12 F

total capacitance = C1 + C2

= 38.2 x 10^-12 F or 38.2 pF .....Ans

Related Questions

Navigate

• Browse (All)
• Browse #
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.