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Chapter 4: Q40E (page 811)

Polystyrene has dielectric constant 2.6 and dielectric strength 2.0×107V/m. A piece of polystyrene is used as a dielectric in a parallel-plate capacitor, filling the volume between the plates. (a) When the electric field between the plates is 80% of the dielectric strength, what is the energy density of the stored energy? (b) When the capacitor is connected to a battery with voltage 500.0 V, the electric field between the plates is 80% of the dielectric strength. What is the area of each plate if the capacitor stores 0.200 mJ of energy under these conditions?

Short Answer

Expert verified
  1. The the energy density of the stored energy when electric field is 80% of the dielectric strength is 2900J/m3.
  2. The area of the plates when the stored energy is 0.200 mJ is 22cm2.

Step by step solution

01

(a) Determination of the energy density of the stored energy when electric field is 80% of the dielectric strength.

The electric field in this case is,

E=0.800Em

The energy density expression is,

u=12εE2 ...(i)

Here, the dielectric strength ε=Kε0

Substitute all the values in equation (i),

u=122.68.854×10-12C2/N·m20.8002.0×107V/m2=2945J/m3≈2900J/m3

So, the energy density stored between the capacitor plates is 2900J/m3.

02

(b) Determination of the area of the plates when the stored energy is 0.200 mJ.

The separation between the plates is,

d=VE=500V0.8002.0×107V/m=3.125×10-5m

Now, the total energy stored is equal to the energy density multiplied by the volume.

U=u×V=uAd

Rearrange and solve for the area A,

A=Uud=0.20×10-32945J/m33.125×10-5=2.2×10-3m2=22cm2

Thus, the area of the plates is 22cm2.

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