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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, $E_m=3脳{10}^{7}V/m$

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

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

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

Step 2

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

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

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

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

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

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

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

$Q=2K{C}_{慰}U$ (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= $11.81脳{10}^{-3}J$

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

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

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

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

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

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

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

Pyrex: K = 4.7, U = $2.20脳{10}^{-3}J$

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

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:

$V=0.5(3脳{10}^{7}V/m)(0.0025)=37500V$

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