# Problem with solution Create a plate capacitor with a certain capacitance

Level 2 (suitable for students)
Level 2 requires school mathematics. Suitable for pupils.

Capacitors are characterized by the electrical capacitance. It tells you how good a capacitor can "store" electric charge. The goal of modern technology is usually to produce the smallest possible capacitors with the largest possible capacitance so that such a component also fits into your smartphone.

You want to have a plate capacitor which has a capacitance of $$C = 0.5 \, \text{nF}$$ (nano farad). The area of a capacitor plate is given as $$A = 12 \, \text{cm}^2$$.

1. How large do you have to choose the distance $$d$$ of the plates to reach this capacitance?
2. What else can you do to achieve the specified capacitance when the plate distance is set to $$d = 1.5 \, \text{mm}$$?
Solution tips

All you need is the formula for the capacitance of a plate capacitor.

## Solution

Solution for (a)

Use:

Formula for the capacitance of a plate capacitor1$C ~=~ \varepsilon_{\text r} \, \varepsilon_{\text 0} \, \frac{A}{d}$

Assuming a vacuum (or air), the relative permittivity is $$\varepsilon_{\text r} \approx 1$$. Insert it and rearrange 1 for the plate distance we are looking for $$d$$:2$d ~=~ \varepsilon_{\text 0} \, \frac{A}{C}$

Substituting the given values into 2, namely $$C = 0.5 \, \text{nF} = 0. 5 \cdot 10^{-9} \, \text{F}$$ and $$A = 12 \, \text{cm}^2 = 12 \cdot 10^{-4} \, \text{m}^2$$ and $$\varepsilon_{\text 0} = 8.8 \cdot 10^{-12} \, \frac{\text{As}}{\text{Vm}}$$:3$d ~=~ 8.8 \cdot 10^{-12} \, \frac{\text{As}}{\text{Vm}} ~\cdot~ \frac{12 \cdot 10^{-4} \, \text{m}^2}{0.5 \cdot 10^{-9} \, \text{F}} ~=~ 2.1 \cdot 10^{-5} \, \text{m} ~=~ 0.021 \, \text{mm}$

As you can see from the result: The plates have to be very close to each other to just reach a small capacitance of $$C = 0.5 \, \text{nF}$$.

Solution for (b)

Since the plate distance $$d = 0.021 \, \text{mm}$$ determined in (a) is incredibly small, the desired capacitance can be achieved in an alternative way without making the distance so small. In this exercise, we want to have at least a spacing of $$d = 1.5 \, \text{mm}$$.

To do this, looking at the formula 1, you must select a suitable dielectric (i.e., a particular material between the capacitor plates), which is characterized by the relative permittivity $$\varepsilon_{\text r}$$.

Rearrange 1 for $$\varepsilon_{\text r}$$:4$\varepsilon_{\text r} = \frac{C \, d}{\varepsilon_{\text 0} \, A}$

Insert the given values ( including the desired distance $$d = 1.5 \, \text{mm} = 1.5 \cdot 10^{-3} \, \text{m}$$):5$\varepsilon_{\text r} ~=~ \frac{0.5 \cdot 10^{-9} \, \text{F} ~\cdot~ 1.5 \cdot 10^{-3} \, \text{m}}{8.8 \cdot 10^{-12} \, \frac{\text{As}}{\text{Vm}} ~\cdot~ 12 \cdot 10^{-4} \, \text{m}^2} ~\approx~ 89$

This is the approximate value of pure water. So you would have to immerse the plate capacitor in pure water to achieve the desired capacitance of $$0.5 \, \text{nF}$$ with plate spacing of $$1.5 \, \text{mm}$$.