Question

Polystyrene has dielectric constant 2.6 and dielectric strength $\mathbf{2}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{\mathbf{7}}\mathbf{}\mathbf{V}\mathbf{/}\mathbf{m}$. 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

1. The the energy density of the stored energy when electric field is 80% of the dielectric strength is $2900\mathrm{J}/{\mathrm{m}}^3$.
2. The area of the plates when the stored energy is 0.200 mJ is $22{\mathrm{cm}}^2$.

Step-by-step solution

(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,

$E=0.800E_m$

The energy density expression is,

$u=\frac{1}{2}\epsilon E^2$ ...(i)

Here, the dielectric strength $\epsilon =K{\epsilon }_0$

Substitute all the values in equation (i),

$$\begin{gathered} u=\frac{1}{2}\left(2.6\right)\left(8.854\times {10}^{-12}{\mathrm{C}}^2/\mathrm{N}·{\mathrm{m}}^2\right){\left[\left(0.800\right)\left(2.0\times {10}^7\mathrm{V}/\mathrm{m}\right)\right]}^2 \\ =2945\mathrm{J}/{\mathrm{m}}^3 \\ \approx 2900\mathrm{J}/{\mathrm{m}}^3 \end{gathered}$$

So, the energy density stored between the capacitor plates is $2900\mathrm{J}/{\mathrm{m}}^3$.

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

The separation between the plates is,

$$\begin{gathered} \mathrm{d}=\frac{\mathrm{V}}{\mathrm{E}} \\ =\frac{\left(500\mathrm{V}\right)}{\left(0.800\right)\left(2.0\times {10}^7\mathrm{V}/\mathrm{m}\right)} \\ =3.125\times {10}^{-5}\mathrm{m} \end{gathered}$$

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

$$\begin{gathered} U=u\times V \\ =uAd \end{gathered}$$

Rearrange and solve for the area A,

$$\begin{gathered} \mathrm{A}=\frac{\mathrm{U}}{\mathrm{ud}} \\ =\frac{\left(0.20\times {10}^{-3}\right)}{\left(2945\mathrm{J}/{\mathrm{m}}^3\right)\left(3.125\times {10}^{-5}\right)} \\ =2.2\times {10}^{-3}{\mathrm{m}}^2 \\ =22{\mathrm{cm}}^2 \end{gathered}$$

Thus, the area of the plates is $22{\mathrm{cm}}^2$.