Question

You have two plates of copper, a sheet of mica ($\mathbf{thickness}\mathbf{}\mathbf{=}\mathbf{}\mathbf{0}\mathbf{.}\mathbf{10}\mathbf{}\mathbf{mm}\mathbf{,}\mathbf{}\mathbf{\kappa }\mathbf{}\mathbf{=}\mathbf{}\mathbf{5}\mathbf{.}\mathbf{4}$), a sheet of glass (), and a slab of paraffin ($\mathbf{thickness}\mathbf{}\mathbf{=}\mathbf{}\mathbf{2}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{mm}\mathbf{,}\mathbf{}\mathbf{\kappa }\mathbf{=}\mathbf{}\mathbf{7}\mathbf{.}\mathbf{0}$). To make a parallel-plate capacitor with the largest C, which sheet should you place between the copper plates?

Short answer

The sheet placed between the copper plates to maximize the capacitance C will be mica sheet.

Step-by-step solution

The given data

• Thickness of mica sheet,$\mathrm{a}=0.10\mathrm{mm}$
• Thickness of glass sheet,$\mathrm{a}=2.0\mathrm{mm}$
• Thickness of slab of paraffin,$\mathrm{a}=1.0\mathrm{cm}$
• Dielectric constant of mica,$\kappa =5.4$
• Dielectric constant of glass,$\kappa =7.0$
• Dielectric constant of paraffin,$\kappa =2.0$

Understanding the concept of the capacitance

A capacitor is a device that can store a charge. The capacity to store the charge is called capacitance. The capacitance can be written in terms of permittivity of the free space, the dielectric medium between the two plates, area of the cross-section of two plates, and the separation between the two plates.

We use the formula of the capacitance of the parallel plate capacitor to calculate the capacitance of the capacitor.

Formulae:

The capacitance between the capacitor plates,$C=\frac{{\epsilon }_{0}kA}{d}$ …(i)

Here, C is capacitance,${\epsilon }_0$ is the permittivity of the free space,A is the area of cross-section, andd is the separation between the plates.

Calculation of the dielectric substance that gives the largest capacitance

We can see that the capacitance of the parallel plate capacitor when dielectric material is placed between the plates will be depended only on the dielectric constant and the plate separation considering equation (i).

$C\alpha \frac{k}{d}$

Hence, to maximize the capacitance of the capacitor, we need to choose such a material whose $\frac{k}{d}$value should be greater.

For mica, the thickness a=0.10 mm and the dielectric constant$k=5.4$

$$\begin{gathered} C\alpha \frac{k}{d-a} \\ =\frac{5.4}{d-0.10mm} \end{gathered}$$

And for glass, the thickness b=2 mm and the dielectric constant $k=7$
$$\begin{gathered} C\alpha \frac{k}{d-a} \\ =\frac{7.0}{d-2mm} \end{gathered}$$

For paraffin, the thickness c=1.0 cm and the dielectric constant k=2

$$\begin{gathered} C\alpha \frac{k}{d-c} \\ =\frac{2}{d-1.0cm} \end{gathered}$$

Hence, from the aboveobservations, we can say that mica will have greater value of $\frac{k}{d}$.

Hence, a mica sheet will be placed to maximize the capacitance.