A parallel-plate capacitor with only air between the plates is charged by connec

Question

A parallel-plate capacitor with only air between the plates is charged by connecting it to a battery. The capacitor is then disconnected from the battery, without any of the charge leaving the plates.A voltmeter reads 50.0 V when placed across the capacitor. When a dielectric is inserted between the plates, completely filling the space, the voltmeter reads 10.0V .
PartA:
What is the dielectric constant of this material?
K=

Part B:
What will the voltmeter read if the dielectric is now pulled partway out so it fills only one-third of the space between the plates?
V= V

Explanation / Answer

a>Q=CV
in both case Q is same which is initialy deposited.
so,C inversely proportional toV
so,C1V1=C2V2
=>C1/C2=10/50
=>C1/C2=1/5
=>C1/C1=1/5 C2=C1

=>=5

b>it si parallel connection of 2/3 rd non diaelectric and 1/3rd dielectric

as C directly proportional to both area between plates and

Ca(2/3rd non dielectric)=C1*2/3

Cb(1/3rd dielectric)=C1/3 *=C1*5/3

net C=Ca+Cb=C1*7/3

again CV=C1V1

=>V=C1*50/(C1*7/3)=21.42 V

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