Question

A cylindrical sandwich cookie has a radius of 0.75 in. The cookie is made from two wafers, each of thickness 0.15 in, which are perfect dielectrics of relative permittivity \(\epsilon_{r}=2.8 .\) Between the wafers is a layer of cream filling of thickness 0.1 in which is a perfect dielectric of relative permittivity \(\epsilon_{r}=2.2 .\) Find the overall capacitance of the cookie. Hint: Capacitances in series combine as \(\frac{1}{\frac{1}{C_{1}}+\frac{1}{C_{2}}}\).

Short answer

The overall capacitance of the cookie is approximately 1.06 pF.

Step-by-step solution

Identify the layers of the sandwich cookie and their parameters

The cookie is made up of three layers: two wafer layers and one cream filling layer in between. Each wafer has a thickness of 0.15 inches, relative permittivity \( \epsilon_{r} = 2.8 \), and the cream layer has a thickness of 0.1 inches, relative permittivity \( \epsilon_{r} = 2.2 \). The radius of the cylindrical cookie is 0.75 inches, which will be used to find the area.

Calculate the area of the capacitor plates

The area \( A \) of each layer (modeled as parallel plates) can be calculated using the formula for the area of a circle, \( A = \pi r^2 \). With a radius \( r = 0.75 \) inches, the area is \( A = \pi (0.75)^2 \approx 1.767 \) square inches.

Calculate the capacitance of each layer

The capacitance \( C \) of a dielectric layer can be determined using the formula \( C = \frac{\epsilon_{r} \epsilon_{0} A}{d} \), where \( \epsilon_{0} = 8.854 \times 10^{-12} \text{ F/m} \) is the permittivity of free space, \( A \) is the area, and \( d \) is the thickness of the dielectric layer. Calculate for each layer:- Wafer layers: \( C_{\text{wafer}} = \frac{2.8 \times 8.854 \times 10^{-12} \times 1.767}{0.15 \times 0.0254} \)- Cream layer: \( C_{\text{cream}} = \frac{2.2 \times 8.854 \times 10^{-12} \times 1.767}{0.1 \times 0.0254} \)Calculate these values.

Convert inches to meters for calculation

To use the values in the calculation for capacitance, convert thickness from inches to meters by multiplying with \( 0.0254 \), the conversion factor from inches to meters. Therefore:- Thickness of wafer \( = 0.15 \times 0.0254 \) meters.- Thickness of cream \( = 0.1 \times 0.0254 \) meters. Use these converted values in Step 3.

Calculate numerical values for capacitance of each layer

By substituting the converted values into the capacitance formula for each layer, perform the calculations using the approximated area from Step 2:- \( C_{\text{wafer}} \approx \frac{2.8 \times 8.854 \times 10^{-12} \times 1.767}{0.15 \times 0.0254} \)- \( C_{\text{cream}} \approx \frac{2.2 \times 8.854 \times 10^{-12} \times 1.767}{0.1 \times 0.0254} \)Simplify and calculate these values.

Combine capacitances in series

Given that the capacitances are in series, use the series combination formula \( \frac{1}{C_{\text{total}}} = \frac{1}{C_{\text{wafer}}} + \frac{1}{C_{\text{cream}}} + \frac{1}{C_{\text{wafer}}} \) to find the total capacitance of the system. Calculate \( C_{\text{total}} \).

Calculate total capacitance

Perform the calculation based on Step 6 to arrive at the total capacitance of the sandwich cookie. Follow the sequence of simplification to find \( C_{\text{total}} \) in Farads.

Key concepts

Cylindrical Capacitors

Cylindrical capacitors can be thought of as a type of capacitor where the conducting surfaces are cylindrical in shape rather than the standard flat plates. In the context of our sandwich cookie example, each layer can be viewed as a separate capacitor having one cylindrical face.
The cookie’s cylindrical shape affects how capacitance is calculated. The capacitance of each cylindrical wafer is found using the formula for capacitance in cylindrical systems. However, for simplicity, we approximate each layer as having parallel plates with effective area governed by the radius of the cookie.
This simplification leverages the geometric similarity to standard flat plate capacitors, understood as discs in this case. Understanding the field distribution within these systems helps us to capture the defining role the geometry holds in overall capacitance.

Dielectric Materials

Dielectric materials are insulating substances positioned between the conducting plates of a capacitor. They are able to store electrical energy while increasing the capacitor’s capacitance compared to air or vacuum.
In the cookie example, both the wafers and the cream filling serve as dielectric materials. They both have their relative permittivity or dielectric constant, which describes their ability to hold an electric field. The wafers have a permittivity of 2.8, while the cream’s permittivity is 2.2.
By comparing these values, it’s clear that the wafer material is better at storing charge than the cream. Choosing dielectric materials with higher permittivity is a way to increase a capacitor’s storage capability without changing its size.

Series Capacitance

When capacitors are placed in a series configuration, their total capacitance is reduced. This is because the charge on the series-connected capacitors must be the same on each, but the potential difference across them is additive.
The formula for series capacitance requires inverting the sum of the inverses of individual capacitances. In our scenario, since three dielectric layers are involved (wafer-cream-wafer), we employ the series capacitance formula:

• \( \frac{1}{C_{\text{total}}} = \frac{1}{C_{\text{wafer}}} + \frac{1}{C_{\text{cream}}} + \frac{1}{C_{\text{wafer}}} \)

This configuration mimics the flow of current through each discrete layer of the cookie, ensuring the overall potential drop is minimized, leading to the correct energy storage solution for this system.

Permittivity

Permittivity represents a material's ability to permit electric field lines to pass through it. It is a core parameter in determining a capacitor's performance, as higher permittivity means better energy storage in capacitors.
This concept is split into two parts: the permittivity of free space, a constant denoted by \( \epsilon_{0} \) and the relative permittivity \( \epsilon_{r} \), which is specific to materials. In our cookie exercise, the different layers have unique permittivities because they involve different materials (wafer vs. cream filling).
The total permittivity \( \epsilon \) for each layer becomes \( \epsilon = \epsilon_{r} \times \epsilon_{0} \). This understanding allows us to effectively calculate the capacitance within each layer via the formula \( C = \frac{\epsilon A}{d} \), highlighting how varied materials affect capacitance.