A parallel-plate capacitor is made from two plates 12.0 cm on each side and 4.40

Question

A parallel-plate capacitor is made from two plates 12.0 cm on each side and 4.40 mm apart. Half of the space between these plates contains only air, but the other half is filled with Plexiglas of dielectric constant 3.45. (See the figure below(Figure 1) .) An 17.0 V battery is connected across the plates.

Part A

What is the capacitance of this combination?

Part B

How much energy is stored in the capacitor?

Part C

If we remove the Plexiglas, but change nothing else, how much energy will be stored in the capacitor?

Explanation / Answer

Work this as two capacitors in parallel. each one has half the area, or an area of 12x6=72 cm² or 72e-4 m²

The air one, C = (8.854e-12)(72e-4) /(0.0040)
The other, C = (8.854e-12)(3.45)(72e-4) /(0.0040)

add the two for the total C

total C 2.14*10^-11

Parallel plate cap
C = (A/d) in Farads
is 8.854e-12 F/m
is dielectric constant (vacuum = 1)
A and d are area of plate in m² and separation in m

The energy is in joules, from

U = 0.5 * (C * V^2).

U = 0.5 * (2.14*10^-11*17^2)

U = 3.09*10^-9 J

Removing the plastic, the capacitance of both sections is the same (air or vacuum), recalculate for that

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