You are designing capacitors for various applications. For one application, you want the maximum possible stored energy. For another, you want the maximum stored charge. For a third application, you want the capacitor to withstand a large applied voltage without dielectric breakdown. You start with an air-filled parallel-plate capacitor that has C0 = 6.00 pF and plate separation of 2.50 mm. You then consider the use of each of the dielectric materials listed in Table 24.2. In each application,

the dielectric will fill the volume between the plates, and the electric field between the plates will be 50% of the dielectric strength given in the table.

(a) For each of the five materials given in the table, calculate the energy stored in the capacitor. Which dielectric allows the maximum stored energy?

(b) For each material, what is the charge Qstored on each plate of the capacitor?

(c) For each material, what is the voltage applied across the capacitor?

(d) Is one dielectric material in the table your best choice for all three applications?

We want to calculate the energy storage of the capacitor. It is related to the capacitance of the capacitor by the equation

$U=\frac{1}{2}K{c}_o{V}^2$ (1)

Where the potential difference equals Ed where E represents 50% of the dielectric strength of each material, equation (1) will be in the form

$U=\frac{1}{2}K{c}_o{\left(Ed\right)}^2$ (2)

Where the term K $C_o$, represents the capacitance after the dielectric material is inserted. Now let us plug our values for K and E for each material to get U of each material where $E=0.50E_m$

Polycarbonate: K = 2.8, $E_m=3\times {10}^{7}V/m$

$$\begin{gathered} U=\frac{1}{2}K{C}_o{\left(0.50E_{m}d\right)}^2 \\ U=\frac{1}{2}\left(2.8\right)\left(6.0\times {10}^{-12}\right){\left(0.50\times 3\times {10}^7\mathrm{V}/\mathrm{m}\times 0.0025\mathrm{m}\right)}^2 \\ U=11.81\times {10}^{-3}\mathrm{J} \\ =6\times {10}^7\mathrm{V}/\mathrm{m} \end{gathered}$$

Polyester: K = 3.3 $E_m=6\times {10}^{7}V/m$,

$$\begin{gathered} U=\frac{1}{2}K{C}_o{\left(0.50E_{m}d\right)}^2 \\ U=\frac{1}{2}\left(2.8\right)\left(6.0\times {10}^{-12}\right){\left(0.50\times 6\times {10}^7\mathrm{V}/\mathrm{m}\times 0.0025\mathrm{m}\right)}^2 \\ U=55.68\times {10}^{-3}\mathrm{J} \end{gathered}$$

Step 2

Polypropylene: K = 2.2, $E_m=7\times {10}^{7}V/m$

$$\begin{gathered} U=\frac{1}{2}K{C}_o{\left(0.50E_{m}d\right)}^2 \\ U=\frac{1}{2}\left(2.8\right)\left(6.0\times {10}^{-12}\right){\left(0.50\times 7\times {10}^7\mathrm{V}/\mathrm{m}\times 0.0025\mathrm{m}\right)}^2 \\ U=50.53\times {10}^{-3}\mathrm{J} \end{gathered}$$

Polystyrene: K =2.6, $E_m=2\times {10}^{7}V/m$

$$\begin{gathered} U=\frac{1}{2}K{C}_o{\left(0.50E_{m}d\right)}^2 \\ U=\frac{1}{2}\left(2.8\right)\left(6.0\times {10}^{-12}\right){\left(0.50\times 2\times {10}^7\mathrm{V}/\mathrm{m}\times 0.0025\mathrm{m}\right)}^2 \\ U=4.87\times {10}^{-3}\mathrm{J} \end{gathered}$$

Pyrex K 4.7, localid="1668315757830" $E_m=2\times {10}^{7}V/m$

$$\begin{gathered} U=\frac{1}{2}K{C}_o{\left(0.50E_{m}d\right)}^2 \\ U=\frac{1}{2}\left(2.8\right)\left(6.0\times {10}^{-12}\right){\left(0.50\times 1\times {10}^7\mathrm{V}/\mathrm{m}\times 0.0025\mathrm{m}\right)}^2 \\ U=2.20\times {10}^{-3}\mathrm{J} \end{gathered}$$

As shown by the results, Polyester has the maximum energy and will be suitable for the first application.

The stored charge in the capacitor is related to the energy storage as shown by the equation

$Q=2K{C}_{o}U$ (3)

Let us put our values for K and U into equation (3) for each material to get the stored charge.

Polycarbonate K = 2.8, U= $11.81\times {10}^{-3}J$

$$\begin{gathered} Q=2K{C}_{o}U \\ Q=2(2.8)(6.0\times {10}^{-12}F)(11.81\times {10}^{-3}J) \\ Q=0.396pC \end{gathered}$$

Polyester: K=3.3, U= $55.68\times {10}^{-3}J$

$$\begin{gathered} Q=2K{C}_{o}U \\ Q=2(3.3)(6.0\times {10}^{-12}F)(55.68\times {10}^{-3}J) \\ Q=2.20pC \end{gathered}$$

Polypropylene: K = 2.2, U =$50.53\times {10}^{-3}J$with the same steps we get

$$\begin{gathered} Q=2K{C}_{o}U \\ Q=2(2.2)(6.0\times {10}^{-12}F)(50.53\times {10}^{-3}J) \\ Q=1.33pC \end{gathered}$$

Polystyrene: K = 2.6, U = $4.87\times {10}^{-3}J$

$$\begin{gathered} Q=2K{C}_{o}U \\ Q=2(2.6)(6.0\times {10}^{-12}F)(4.87\times {10}^{-3}J) \\ Q=0.15pC \end{gathered}$$

Pyrex: K = 4.7, U = $2.20\times {10}^{-3}J$

$$\begin{gathered} Q=2K{C}_{o}U \\ Q=2(4.7)(6.0\times {10}^{-12}F)(2.20\times {10}^{-3}J) \\ Q=0.12pC \end{gathered}$$

As shown by the results, Polyester has the maximum charge and will be suitable for the second application.

The potential difference across the capacitor depends on the electric field between the two plates and it is given by

$$\begin{gathered} V=Ed \\ V=0.5E_{m}d \end{gathered}$$ (4)

Now we can put our values for d and $E_m$into equation (4) for each material to get V

Polycarbonate:

$V=0.5(3\times {10}^{7}V/m)(0.0025)=37500V$

With the same steps, we can get the next for each material

With the same steps, we can get the next for each material

Polyester V= 75000 V

Polypropylene V=87500 V

Polystyrene V= 25000

Pyrex V=12500 V