/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 66 A fuel gauge uses a capacitor to... [FREE SOLUTION] | 91Ó°ÊÓ

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A fuel gauge uses a capacitor to determine the height of the fuel in a tank. The effective dielectric con- stant \(K_{\text {eff }}\) changes from a value of 1 when the tank is empty to a value of \(K\), the dielectric constant of the fuel, when the tank is full. The appropriate electronic circuitry can determine the effective dielectric constant of the combined air and fuel between the capacitor plates. Each of the two rectangular plates has a width \(w\) and a length \(L\) (Fig. \(\mathbf{P 2 4 . 6 6}\) ). The height of the fuel between the plates is \(h\). You can ignore any fringing effects. (a) Derive an expression for \(K_{\text {eff }}\) as a function of \(h\). (b) What is the effective dielectric constant for a tank \(\frac{1}{4}\) full, \(\frac{1}{2}\) full, and \(\frac{3}{4}\) full if the fuel is gasoline \((K=1.95) ?\) (c) Repeat part (b) for methanol \((K=33.0)\). (d) For which fuel is this fuel gauge more practical?

Short Answer

Expert verified
The short answer will depend on the calculations made in Steps 2 and 3, and the comparison made in Step 4.

Step by step solution

01

Derive expression for \(K_{\text {eff }}\)

Let the distance between the plates be \(d\). The effective dielectric constant \(K_{\text {eff }}\) can be derived by considering the capacitor as two capacitors in series, one filled with air and the other with fuel. The dielectric constant of air is 1, and the height of the air part is \(d-h\). The dielectric constant of fuel is \(K\) and the height is \(h\). The total capacitance of the capacitor would be \(C_{total}=(\varepsilon_{0}wL/(d-h\))\(C_{2}=\varepsilon_{0}KwL/h\), where \(C_{1}\) and \(C_{2}\) are the capacitances of the air and fuel parts respectively. The total capacitance when two capacitors are in series is given by \(1/C_{total}=1/C_{1}+1/C_{2}\). Solve this equation for \(K_{\text {eff }}=\varepsilon_{0}K_{eff}wL/d\), where \(K_{eff}\) is the effective dielectric constant.
02

Calculate \(K_{\text {eff }}\) for different fuel heights

To find \(K_{\text {eff }}\) for a tank \(\frac{1}{4}\) full, \(\frac{1}{2}\) full, and \(\frac{3}{4}\) full for gasoline, substitute the corresponding \(h\) values and \(K = 1.95\) into the expression derived in Step 1 and calculate \(K_{\text {eff }}\) respectively.
03

Repeat calculations for methanol

Repeat the calculations in Step 2, but this time using \(K = 33.0\), the dielectric constant for methanol.
04

Determine the more practical fuel

The fuel gauge will be more practical for the type of fuel that leads to larger changes in the effective dielectric constant for changes in fuel height. Compare the results from Steps 2 and 3 to make this determination.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Dielectric Constant
The dielectric constant is a measure of a material's ability to store charge and affects how capacitors behave when different substances are between the plates. In this problem, the dielectric constant varies based on the fuel level in a tank, impacting the capacitance measurement used by a fuel gauge.
When a tank is empty, the dielectric constant of air, which is 1, dominates. As fuel fills the tank, the dielectric constant increases to the value characteristic of the fuel used. For gasoline, the constant is 1.95 and for methanol, it's 33.0. The effective dielectric constant, denoted as \(K_{\text{eff}}\), changes between these values as the fuel level changes.
  • If the tank is full, \(K_{\text{eff}}\) equals the dielectric constant of the fuel.
  • When the tank is empty, \(K_{\text{eff}}\) is 1, as only air is between the capacitor plates.
  • For intermediate levels, \(K_{\text{eff}}\) can be calculated by considering the tank as comprising two series capacitors—one with air and the other with fuel.
Understanding how \(K_{\text{eff}}\) changes helps engineers design more accurate fuel gauges.
Fuel Gauge Mechanics
Fuel gauges often utilize capacitors, which change capacitance based on the material between their plates. This feature is employed in determining the fuel level in a tank. Here's how it works:
First, the capacitor plates are installed in the fuel tank. As the fuel level changes, the material between the plates changes from air to fuel, significantly affecting capacitance.
It's important to accurately establish the effective dielectric constant at different fuel levels. This helps electronic circuits measure fuel height based on the changing capacitance. By detecting changes in \(K_{\text{eff}}\), the fuel gauge can effectively calculate the fuel level, whether the tank is 1/4, 1/2, or 3/4 full.
  • For instance, when the fuel is gasoline, the gauge must measure changes from 1 to 1.95.
  • For methanol, this range is more substantial, from 1 to 33.0.
This technique allows for precise fuel level readings by considering the different dielectric constant values between air and fuel.
Capacitor Design
Capacitor design is critical in developing a reliable fuel gauge that can measure fuel levels accurately. The design considers multiple factors to optimize the capacitor's performance in a fuel gauge.
The main challenge is to handle capacitors" changes in capacitance as they operate with a combination of air and fuel. The approach involves treating the setup as two separate capacitors in series. Here's a simplified process:
  • Design plates considering the tank's dimensions, such as width \(w\) and length \(L\).
  • Calculate \(K_{\text{eff}}\) using the derivation of effective capacitance when combining air-filled and fuel-filled capacitors.
  • Ensure the design allows accurate measurements regardless of fuel level.
The goal is to create a system sensitive enough to distinguish small changes in \(K_{\text{eff}}\). Thus, a tank that is partially filled tweaks the capacitor composition, adjusting the fuel gauge's reading. Understanding the design principles allows for choosing the correct plate size and spacing, improving accuracy, and providing a reliable fuel measurement system.

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Most popular questions from this chapter

DATA Your electronics company has several identical capacitors with capacitance \(C_{1}\) and several others with capacitance \(C_{2}\). You must determine the values of \(C_{1}\) and \(C_{2}\) but don't have access to \(C_{1}\) and \(C_{2}\) individually. Instead, you have a network with \(C_{1}\) and \(C_{2}\) connected in series and a network with \(C_{1}\) and \(C_{2}\) connected in parallel. You have a \(200.0 \mathrm{~V}\) battery and instrumentation that measures the total \(\mathrm{en}\) ergy supplied by the battery when it is connected to the network. When the parallel combination is connected to the battery, \(0.180 \mathrm{~J}\) of energyis stored in the network. When the series combination is connected. \(0.0400 \mathrm{~J}\) of energy is stored. You are told that \(C_{1}\) is greater than \(C_{2}\) (a) Calculate \(C_{1}\) and \(C_{2}\). (b) For the series combination, does \(C_{1}\) or \(C_{2}\) store more charge, or are the values equal? Does \(C_{1}\) or \(C_{2}\) store more energy, or are the values equal? (c) Repeat part (b) for the parallel combination.

After combing your hair on a dry day, some of your hair stands up, forced away from your head by electrostatic repulsion. (a) Estimate the length \(L\) of your hair. (b) Using the average linear mass density of hair, which is \(65 \mu \mathrm{g} / \mathrm{cm},\) estimate the mass \(m\) of one of your hairs. (c) Estimate the number \(N\) of hairs that stand after combing. (d) Assume that the comb has taken away a charge \(-2 Q,\) and that your hair has therefore gained an amount of charge \(2 Q .\) Assume further that half of this charge resides next to your head and the other half is distributed at the ends of the \(N\) strands that stand up. Assume the electrostatic force that lifted a hair was twice its weight. Show that this leads to \(2 m g=\frac{1}{4 \pi \epsilon_{0}} \frac{Q^{2} / N}{L^{2}}\) Use this equation to estimate the charge \(Q\) that resides on your head. (e) If your head were a sphere with radius \(R\), it would have a capacitance of \(4 \pi \epsilon_{0} R .\) Estimate the radius of your head; then use your estimate to determine your head's capacitance. (f) Use your result to estimate the potential attained due to combing. (The surprising result illustrates an interesting feature of static electricity.)

A parallel-plate capacitor is made from two plates \(12.0 \mathrm{~cm}\) on each side and \(4.50 \mathrm{~mm}\) apart. Half of the space between these plates contains only air, but the other half is filled with Plexiglas of dielectric constant 3.40 (Fig. \(\mathbf{P} 24.64)\). An \(18.0 \mathrm{~V}\) battery is connected across the plates. (a) What is the capacitance of this combination? (Hint: Can you think of this capacitor as equivalent to two capacitors in parallel?) (b) How much energy is stored in the capacitor? (c) If we remove the Plexiglas but change nothing else, how much energy will be stored in the capacitor?

You have two identical capacitors and an external potential source. (a) Compare the total energy stored in the capacitors when they are connected to the applied potential in series and in parallel. (b) Compare the maximum amount of charge stored in each case. (c) Energy storage in a capacitor can be limited by the maximum electric field between the plates. What is the ratio of the electric field for the series and parallel combinations?

\- Polystyrene has dielectric constant 2.6 and dielectric strength \(2.0 \times 10^{7} \mathrm{~V} / \mathrm{m} .\) A piece of polystyrene is used as a dielectric in a parallel-plate capacitor, filling the volume between the plates. (a) When the electric field between the plates is \(80 \%\) of the dielectric strength, what is the energy density of the stored energy? (b) When the capacitor is connected to a battery with voltage \(500.0 \mathrm{~V},\) the electric field between the plates is \(80 \%\) of the dielectric strength. What is the area of each plate if the capacitor stores \(0.200 \mathrm{~mJ}\) of energy under these conditions?

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