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.)

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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 5
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.)
Transcribed Image Text:QUESTION 5 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.)
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.
Transcribed Image Text: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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