Question

Design a parallel plate capacitor with a capacitance of \(47.0 \mathrm{pF}\) and a capacity of \(7.50 \mathrm{nC}\). You have available conducting plates, which can be cut to any size, and Plexiglas sheets, which can be cut to any size and machined to any thickness. Plexiglas has a dielectric constant of 3.40 and a dielectric strength of \(4.00 \cdot 10^{7} \mathrm{~V} / \mathrm{m}\). You must make your capacitor as compact as possible. Specify all relevant dimensions. Ignore any fringe field at the edges of the capacitor plates.

Short answer

The smallest possible size for this parallel plate capacitor has conducting plates with an area of 2.00 x 10^-7 m^2 and a distance of 3.99 x 10^-6 m between them.

Step-by-step solution

Calculate the product of the vacuum and dielectric constants

We can multiply the vacuum permittivity (\(\varepsilon_0\)) with the dielectric constant of Plexiglas (\(\varepsilon_r\)) to get the product of both values: Product = \(\varepsilon_0 \varepsilon_r = (8.854 \times 10^{-12} \mathrm{F} / \mathrm{m}) \times 3.40\)

Calculate the distance d between the capacitor plates

We can use the given dielectric strength to calculate the minimum distance between the plates while maintaining safety. Since the dielectric strength is expressed as \(V/m\), we need to consider the capacity of the capacitor (\(7.50 \mathrm{nC}\)) and convert it to the voltage value by dividing the capacity by the given capacitance: \(V = \frac{Q}{C} = \frac{7.50 \times 10^{-9} \mathrm{C}}{47.0 \times 10^{-12} \mathrm{F}} = 159.6 \mathrm{V}\) Now, we can use this voltage value and the given dielectric strength to calculate the minimum distance between the plates: \(d = \frac{V}{E} = \frac{159.6 \mathrm{V}}{4.00 \times 10^7 \mathrm{V} / \mathrm{m}} = 3.99 \times 10^{-6} \mathrm{m}\)

Calculate the Area of the Capacitor Plates

We can now use the given capacitance value, calculated product of permittivity and distance (\(d\)) to find the area of the conducting plates: \(A = \frac{C \cdot d}{\varepsilon_0 \varepsilon_r} = \frac{47.0 \times 10^{-12} \mathrm{F} \cdot 3.99 \times 10^{-6} \mathrm{m}}{8.854 \times 10^{-12} \mathrm{F} / \mathrm{m} \cdot 3.40} = 2.00 \times 10^{-7} \mathrm{m^2}\)

Final Design

The designed parallel plate capacitor with the smallest possible size has conducting plates with an area of \(2.00 \times 10^{-7} \mathrm{m^2}\) and a distance of \(3.99 \times 10^{-6} \mathrm{m}\) between them. Plexiglas sheets of this area and thickness should be used as the dielectric material.

Key concepts

Parallel Plate Capacitor

A parallel plate capacitor is one of the simplest types of capacitors and is widely used in electrical circuits. It consists of two flat and parallel conducting plates separated by a dielectric material. The capacitance of a parallel plate capacitor depends on three factors:

• The surface area of the plates (\(A\)
• The distance between the plates (\(d\)
• The permittivity of the dielectric material inserted between them (\(\varepsilon\)

The formula to calculate the capacitance (\(C\) is given by:\[C = \frac{\varepsilon A}{d}\]Here, the dielectric material plays a crucial role in increasing the capacitance by reducing the electric field within the capacitor.

Dielectric Constant

The dielectric constant, often symbolized as \(\varepsilon_r\) , is a measure of a material's ability to store electrical energy in an electric field. It is a dimensionless quantity that indicates how much the dielectric material can increase the capacitance of a capacitor compared to a vacuum.

• A higher dielectric constant means a higher capacitance, which is beneficial for making compact capacitors.
• In the exercise, Plexiglas was used as a dielectric with a constant of 3.40, which helps to increase the capacitance of the capacitor.

The dielectric constant is crucial when designing a capacitor, as it determines the effectiveness of the dielectric material in storing electrical energy.

Dielectric Strength

Dielectric strength is the maximum electric field that a dielectric material can withstand without breaking down or losing its insulating properties. It is measured in volts per meter (\(V/m\) and represents the material's ability to resist electrical breakdown.

• In the given problem, Plexiglas has a dielectric strength of \(4.00 \times 10^7 \)
\(V/m\), indicating it can withstand this level of electric field safely. This value determines how thin the dielectric layer can be, while ensuring it doesn't fail.
• A high dielectric strength allows for very thin dielectric layers, making the capacitor smaller but still safe to use.

In designing capacitors, maintaining a balance between dielectric constant and dielectric strength is crucial for efficient performance.

Vacuum Permittivity

Vacuum permittivity, denoted as \(\varepsilon_0\), is a fundamental physical constant that describes the ability of the vacuum to permit electric field lines. It is used as a base for measuring the permittivity of various materials.

• Its value is approximately \(8.854 \times 10^{-12} \)
\(F/m\), and it helps in calculating the capacitance of capacitors along with the material's dielectric constant.
• In capacitor design, any increase in permittivity—due to the use of a dielectric material compared to a vacuum—enhances the capacitance.

Vacuum permittivity is a key element in the formula for capacitance and is always combined with the dielectric constant to calibrate the effectiveness of the dielectric material in a capacitor design.