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When a certain air-fillled parallel-plate ca-
pacitor is connected across a battery, it ac-
quires a charge (on each plate) of magnitude
95 microC. While the battery connection is main-
tained, a dielectric slab is inserted into the
space between the capacitor plates and com-
pletely fills this region. This results in the ac-
cumulation of an additional charge of 328 microC
on each plate.
What is the dielectric constant of the slab?

2007-03-23 05:10:30 · 1 answers · asked by kpm0729 1 in Science & Mathematics Physics

1 answers

Q = CV

C1 = Q1/V (air filled capacitor) Q1 = 95x10^-6

C2 = Q2/V where Q2 = Q1 + 328 x10^-6

V is constant throughout, therefore

Q1/C1 = Q2/C2

C2/C1 = Q2/Q1

The ratio of the capacitors simplifies to epsilon2/epsilon1

Hence epsilon2 = (Q2/Q1)*epsilon1

2007-03-23 05:17:19 · answer #1 · answered by dudara 4 · 0 0

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