Problem with solution: Create a Plate Capacitor with a Certain Capacitance

Plate codensator with area \( A \), charge \( Q \), distance \( d \), voltage \( U \) and the dielectric between the plates.

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 \( A = 12 \, \text{cm}^2 \).

How large do you have to choose the distance \( d \) of the plates to reach this capacitance?
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 for (a)

$$ \begin{align} C ~=~ \varepsilon_{\text r} \, \varepsilon_{\text 0} \, \frac{A}{d} \end{align} $$

$$ \begin{align} d = \varepsilon_{\text 0} \, \frac{A}{C} \end{align} $$

$$ \begin{align} d ~&=~ 8.8 \cdot 10^{-12} \, \frac{\mathrm{As}}{\mathrm{Vm}} ~\cdot~ \frac{12 \cdot 10^{-4} \, \mathrm{m}^2}{0.5 \cdot 10^{-9} \, \mathrm{F}} \\\\
~&=~ 2.1 \cdot 10^{-5} \, \text{m} ~=~ 0.021 \, \mathrm{mm} \end{align} $$

Solution for (b)

$$ \begin{align} \varepsilon_{\text r} = \frac{C \, d}{\varepsilon_{\text 0} \, A} \end{align} $$

$$ \begin{align} \varepsilon_{\text r} ~&=~ \frac{0.5 \cdot 10^{-9} \, \text{F} ~\cdot~ 1.5 \cdot 10^{-3} \, \text{m}}{8.8 \cdot 10^{-12} \, \frac{\text{As}}{\text{Vm}} ~\cdot~ 12 \cdot 10^{-4} \, \text{m}^2} \\\\
~&\approx~ 89 \end{align} $$

More exercises with solutions

Charge in a thundercloud

The exercise teaches you how to calculate the charge of a thundercloud using the capacitance and E-field in the plate capacitor.

Capacitor with and without dielectric in comparison

In this exercise with solution about the plate capacitor, the capacitance, voltage and energy must be compared with and without dielectric.

Plate Capacitor: Voltage, Capacitance and Eletric Force

Here the plate capacitor is simply explained. With the help of the capacitor you will learn about the electric voltage, electric field and electric capacitance and how to calculate them for a plate capacitor.

Derivations & Experiments

Plate Capacitor: Potential, E-field, Charge and Capacitance

Here, the electrostatic potential, electric field and the capacitance of a plate capacitor are derived using Laplace's equation.

Related formulas

Plate Capacitor (Force, Voltage, Charge, Distance)

$$ F ~=~ \class{red}{q} \, \frac{U}{d} $$

Plate Capacitor (Capacitance)

$$ C ~=~ \varepsilon_0 \, \varepsilon_{\text r} \, \frac{A}{d} $$

Plate Capacitor (Energy, Electric Field)

$$ W_{\text{e}} ~=~ \frac{1}{2} \, \varepsilon_0 \, \varepsilon_{\text r} \, V \, \class{purple}{E}^2 $$

Plate Capacitor (Electric Field, Voltage, Distance)

$$ \class{purple}{E} ~=~ \frac{U}{d} $$

Capacitor (Energy, Capacitance, Charge)

$$ W_{\text{e}} ~=~ \frac{1}{2} \, \frac{Q^2}{C} $$

Capacitor (Energy, Voltage, Capacitance)

$$ W_{\text e} ~=~ \frac{1}{2} \, C \, U^2 $$

Plate capacitor (potential, voltage, distance)

$$ \varphi_x = - \frac{U}{d} \, x ~+~ \varphi_1 $$

Plate Capacitor (Attraction Force, Voltage)

$$ F ~=~ \frac{\varepsilon_0 \, \varepsilon_{\text r} \, A}{2d^2}\, U^2 $$

Plate Capacitor (Attraction Force, Charge)

$$ F ~=~ \frac{Q^2}{2\varepsilon_0 \, \varepsilon_{\text r} A} $$

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