Find the equivalent capacitance in pF for the dielectric capacitor of problem 49 using values of κ1 = 10.7 and κ2 = 12.0. (Answer in 5 sig. figs.) QUESTION 5Find the equivalent capacitance in pF for thedielectric capacitor of problem 25.49 using valuesof K₁ = 10.7 and K2 = 12.0. (5 sig. figs.) 49 Figure 25-48 shows a parallel-plate ca-pacitor with a plate area A = 7.89 cm² andplate separation d = 4.62 mm. The top half ofthe gap is filled with material of dielectricconstant K₁ 11.0; the bottom half is filledwith material of dielectric constant K₂What is the capacitance?=12.0.=K2K1Figure 25-48Problem 49.

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Answered: Find the equivalent capacitance in pF…

Find the equivalent capacitance in pF for the dielectric capacitor of problem 25.49 using values of K₁ = 10.7 and K2= 12.0. (5 sig. figs.)

Glencoe Physics: Principles and Problems, Student Edition

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ISBN:9780078807213

Author:Paul W. Zitzewitz

Publisher:Paul W. Zitzewitz

Chapter21: Electric Fields

Section: Chapter Questions

Problem 82A

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Combination of Capacitors

The connection of capacitors can be established in a circuit in two ways.

Question

Find the equivalent capacitance in pF for the dielectric capacitor of problem 49 using values of κ1 = 10.7 and κ2 = 12.0. (Answer in 5 sig. figs.)

49 Figure 25-48 shows a parallel-plate ca-
pacitor with a plate area A = 7.89 cm² and
plate separation d = 4.62 mm. The top half of
the gap is filled with material of dielectric
constant K₁ 11.0; the bottom half is filled
with material of dielectric constant K₂
What is the capacitance?
=
12.0.
=
K2
K1
Figure 25-48
Problem 49.

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