Given a 7.4 pF air filled capacitor, you are asked to convert it to a capacitor that can store up to $\mathbf{7}\mathbf{.}\mathbf{4}\mathbf{\mu J}$with a maximum potential difference of 652 v. Which dielectric in table should you use to fill the gap in the capacitor if you do not allow for a margin of error?

a) Capacitance value without dielectric, ${\mathrm{C}}_0=7.4\times {10}^{-12}\mathrm{F}$

b) Maximum potential difference, V = 652 V

c) Energy stored in capacitor with dielectric, $U=7.4\times {10}^{-6}J$

If the space between the plates of a capacitor is completely filled with a dielectric material, the capacitance is C increased by a factor k, called the dielectric constant, which is characteristic of the material. In a region that is completely filled by a dielectric, all electrostatic equations containing${\mathit{\epsilon }}_{\mathbf{0}}$ must be modified by replacing${\mathit{\epsilon }}_{\mathbf{0}}$ with$\mathit{k}{\mathit{\epsilon }}_{\mathbf{0}}$.

We can use the relation between capacitance without dielectric and with a dielectric material and the equation for energy in terms of capacitance and potential difference to find the dielectric constant. Using the product of the capacitance of the capacitor without dielectric (or with air, ) and the dielectric constant of the material, we can get the value of the capacitance with dielectric. Thus, by substituting this in the energy equation, we can find the value of the dielectric constant.

Formulae:

The capacitance relation with dielectric to without dielectric,$C_k=k\times C_0$ …(i)

Here,$C_k$ is capacitance with dielectric, k is dielectric constant,$C_0$ is capacitance without dielectric.

The energy stored between the capacitors,$U=\frac{1}{2}\times C_k{V}^2$ …(ii)

Here, U is the stored energy,$C_k$ is the capacitance with dielectric and V is the potential difference between two plates.

Using the given data in the substituted equation of (i) in equation (ii), we can get the value of the dielectric constant as follows:

$$\begin{gathered} C_0\times k=\frac{2U}{V^2} \\ k=\frac{2U}{C_0{V}^2} \\ =\frac{2\times 7.4\times {10}^{-6}J}{7.4\times {10}^{-12}F\times {\left(652V\right)}^2} \\ =4.7 \end{gathered}$$

Hence, comparing this dielectric value, we get that the material is Pyrex.