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 \(3.00 \cdot 10^{7} \mathrm{~V} / \mathrm{m}\). 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

Question: Design a parallel plate capacitor with a capacitance of 47.0 pF and a capacity of 7.50 nC using Plexiglas as the dielectric material. Make sure the capacitor is as compact as possible. Provide the final dimensions and specifications for the designed capacitor. Answer: To design the compact parallel plate capacitor with a capacitance of 47.0 pF and a capacity of 7.50 nC using Plexiglas, first find the area of the plates, the minimum thickness of the insulator between them, and then the specific dimensions of the plates. The final dimensions and specifications for the designed capacitor are as follows: - Area of the plates: *insert calculated area here* - Minimum thickness of the insulator: *insert calculated d_min here* - Length and width of the plates: *insert length and width here* Note: Please follow the step-by-step solution above to obtain the correct values for the final dimensions and specifications.

Step-by-step solution

Find the area of the capacitor plates

Using the capacitance equation: \(C = \epsilon_{0} \epsilon_{r} \frac{A}{d}\), where \(C\) is the capacitance, \(\epsilon_{0}\) is the vacuum permittivity \((8.854\times10^{-12} \mathrm{F} / \mathrm{m})\), \(\epsilon_{r}\) is the dielectric constant of Plexiglas, \(A\) is the area of the capacitor plates, and \(d\) is the distance between the plates. Given the capacitance \(C = 47.0 \mathrm{pF}\) and the dielectric constant of Plexiglas \(\epsilon_{r} = 3.40\), we can rearrange the equation to find the area of the plates: \(A = \frac{C \cdot d}{\epsilon_{0} \epsilon_{r}}\)

Determine the minimum thickness between capacitor plates

The dielectric strength of Plexiglas is \(3.00 \times 10^{7} \mathrm{V} / \mathrm{m}\). We can use the capacity (\(Q = 7.50 \mathrm{nC}\)) and the dielectric strength to find the voltage \(V\) across the capacitor: \(V = \frac{Q}{C}\) Now, we can find the minimum thickness \(d\) based on the dielectric strength and calculated voltage: \(d_{min} = \frac{V}{E_{D}} \)

Calculate the area and dimensions of the capacitor plates

Now that we have the minimum thickness between the plates, we can return to the area calculation from step 1 and find \(A\). Knowing the area, we can then choose the dimensions (length and width) of the plates to design a compact parallel plate capacitor.

Provide final dimensions and specifications

After calculating the area and dimensions of the capacitor plates, as well as the thickness of the insulator separating them, we can list the final dimensions and specifications for the designed parallel plate capacitor.

Key concepts

Capacitance Calculation

Understanding how to calculate the capacitance of a parallel plate capacitor is crucial for its design. The formula used is:

\(C = \epsilon_{0} \epsilon_{r} \frac{A}{d}\)

where:

• \(C\) is the capacitance of the capacitor;
• \(\epsilon_{0}\) is the vacuum permittivity (approximately \(8.854\times10^{-12} \mathrm{F/m}\));
• \(\epsilon_{r}\) is the dielectric constant, a material-specific value;
• \(A\) is the area of the capacitor plates; and
• \(d\) is the distance between the plates.

In layman's terms, the capacitance value reflects how much electrical charge the capacitor can store. To achieve the desired capacitance, one can adjust parameters like the area of the plates and the distance between them, knowing the properties of the material in between the plates (the dielectric).

If a capacitor needs to be compact, it's crucial to optimize the area and the separation distance of the plates without compromising the desired capacitance. Using materials with higher dielectric constants can allow for smaller capacitors as it increases \(C\) without increasing \(A\) or decreasing \(d\).

Dielectric Constant

The dielectric constant, denoted as \(\epsilon_{r}\), is an essential property of insulating materials. Physically, \(\epsilon_{r}\) represents the ability of a material to polarize in response to an applied electric field, hence storing more electric charge. It's a dimensionless quantity that compares the permittivity of a material to the permittivity of a vacuum.

• A higher \(\epsilon_{r}\) means a higher capacitance for the same plate area and separation.
• In the context of the textbook problem, Plexiglas has a dielectric constant of 3.40, which is then multiplied with vacuum permittivity (\(\epsilon_{0}\)) to determine how much the material increases the capacitor's ability to store charge versus a vacuum.

Choosing a material with a high dielectric constant is particularly beneficial when space is a limiting factor and the design requires the capacitor to be as compact as possible, as is the case in the original exercise.

Dielectric Strength

The dielectric strength of a material is the maximum electric field that a material can withstand without breaking down, usually measured in volts per meter (V/m). Breakdown occurs when the electric field is strong enough to start a cascade of electron collisions, making an insulating material conductive, which in the context of capacitors, would lead to a failure.

The problem provides the dielectric strength of Plexiglas (\(3.00 \times 10^{7} \mathrm{V/m}\)). This tells us the maximum voltage that can be applied per meter thickness of Plexiglas before it breaks down. Hence, with the capacitor's charge capacity known, the minimum separation distance between the plates can be determined to ensure operation within safe limits.

• To design a capacitor that will not fail under normal operating conditions, the voltage across the capacitor must not exceed Plexiglas's dielectric strength.
• This will affect the minimum thickness of the plate separation, balancing compact design with safety and efficiency.

It is vital to keep these considerations in mind when determining the dimensions of a capacitor in order to design a device that is not only compact but also reliable and safe for its intended application.