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Chapter 24: Problem 37

The dielectric to be used in a parallel-plate capacitor has a dielectric constant of 3.60 and a dielectric strength of 1.60 \(\times\) 10\(^7\) V>m. The capacitor is to have a capacitance of 1.25 \(\times\) 10\(^{-9}\) F and must be able to withstand a maximum potential difference of 5500 V. What is the minimum area the plates of the capacitor may have?

Short Answer

Expert verified
The minimum plate area needed is approximately 0.0136 m虏.

Step by step solution

01

Understand the Problem

We need to find the minimum area of the plates of a parallel-plate capacitor given specific electrical properties: dielectric constant, dielectric strength, capacitance, and maximum potential difference.
02

Use the Capacitance Formula

Recall the formula for the capacitance of a parallel-plate capacitor: \[ C = \frac{\varepsilon_0 \varepsilon_r A}{d} \]where \(\varepsilon_0\) is the permittivity of free space (\(8.85 \times 10^{-12} \text{ F/m}\)), \(\varepsilon_r\) is the relative permittivity (3.60), \(A\) is the area, and \(d\) is the distance between the plates. Here, \(C = 1.25 \times 10^{-9} \text{ F}\).
03

Apply Dielectric Strength Condition

The dielectric strength provides the maximum electric field (\(E_{max}\)) the material can withstand, given by \(E_{max} = 1.60 \times 10^7 \text{ V/m}\). The potential difference \(V\) is related to the electric field by \(V = E_{max} \cdot d\). Solve for \(d\):\[ d = \frac{V}{E_{max}} = \frac{5500 \, \text{V}}{1.60 \times 10^7 \, \text{V/m}} \approx 3.44 \times 10^{-4} \, \text{m}. \]
04

Rearrange the Capacitance Formula for Area

We know \(C\), \(\varepsilon_0\), \(\varepsilon_r\), and \(d\), and we can solve for \(A\):\[ A = \frac{C \cdot d}{\varepsilon_0 \varepsilon_r} = \frac{1.25 \times 10^{-9} \, \text{F} \cdot 3.44 \times 10^{-4} \, \text{m}}{8.85 \times 10^{-12} \, \text{F/m} \cdot 3.60} \approx 1.36 \times 10^{-2} \, \text{m}^2. \]
05

Solution Conclusion

The calculation yields the minimum area of the capacitor plates necessary to meet the given requirements.

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Key Concepts

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

Dielectric Strength
Dielectric strength is a critical property of insulating materials, especially ones used in capacitors. It refers to the maximum electric field that a material can withstand without breaking down or losing its insulating properties. In simple terms, it's the material's ability to resist an electric breakdown.
For a dielectric material, the dielectric strength is measured in volts per meter (V/m). A higher dielectric strength means that the material can endure a stronger electric field. This is essential in preventing breakdown in capacitors, ensuring the device functions correctly under high voltages.
In the context of capacitor design, knowing the dielectric strength helps determine how thin the material can be without risking failure. Choosing a material with a suitable dielectric strength ensures that the capacitor can operate safely at the intended voltage, as calculated in our exercise.
Parallel-Plate Capacitor
A parallel-plate capacitor is a straightforward type of capacitor that features two conducting plates separated by an insulating material called a dielectric. This setup is widely used due to its simplicity and effectiveness.
The capacitance of this kind of capacitor is determined by the area of the plates, the separation distance between the plates, and the properties of the dielectric in between. Larger plate areas and smaller separation distances increase capacitance, making for a more efficient energy storage device.
Parallel-plate capacitors are commonly used in electronic circuits and can be found in devices ranging from radios to laptops. Understanding the factors that influence their design, like the dielectric material and plate area, is key to optimizing their performance for various applications.
Capacitance Formula
The capacitance formula for a parallel-plate capacitor is: \[ C = \frac{\varepsilon_0 \varepsilon_r A}{d} \]where:
  • \(C\) is the capacitance in farads (F).
  • \(\varepsilon_0\) is the permittivity of free space (\(8.85 \times 10^{-12} \) F/m).
  • \(\varepsilon_r\) is the relative permittivity or dielectric constant of the material.
  • \(A\) is the area of the plates.
  • \(d\) is the separation distance between the plates.
This formula calculates how much electric charge a capacitor can store. The greater the capacitance, the more charge it can hold for a given voltage.
By rearranging the formula, we can find different variables, such as the minimum area of the plates required for a specific capacitance, as done in the exercise. This is crucial for designing capacitors that meet precise electrical criteria.

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Most popular questions from this chapter

A parallel-plate air capacitor is to store charge of magnitude 240.0 pC on each plate when the potential difference between the plates is 42.0 V. (a) If the area of each plate is 6.80 cm\(^2\), what is the separation between the plates? (b) If the separation between the two plates is double the value calculated in part (a), what potential difference is required for the capacitor to store charge of magnitude 240.0 pC on each plate?

Electronic flash units for cameras contain a capacitor for storing the energy used to produce the flash. In one such unit, the flash lasts for \({1 \over 675}\) s with an average light power output of 2.70 \(\times\) 10\(^5\) W. (a) If the conversion of electrical energy to light is 95% efficient (the rest of the energy goes to thermal energy), how much energy must be stored in the capacitor for one flash? (b) The capacitor has a potential difference between its plates of 125 V when the stored energy equals the value calculated in part (a). What is the capacitance?

Some cell walls in the human body have a layer of negative charge on the inside surface and a layer of positive charge of equal magnitude on the outside surface. Suppose that the charge density on either surface is \(\pm\) 0.50 \(\times\) 10\(^{-3}\) C/m\(^2\), the cell wall is 5.0 nm thick, and the cell- wall material is air. (a) Find the magnitude of \(\vec{E}\) S in the wall between the two layers of charge. (b) Find the potential difference between the inside and the outside of the cell. Which is at the higher potential? (c) A typical cell in the human body has a volume of 10\(^{-16}\) m\(^3\). Estimate the total electric-field energy stored in the wall of a cell of this size. (\(Hint\): Assume that the cell is spherical, and calculate the volume of the cell wall.) (d) In reality, the cell wall is made up, not of air, but of tissue with a dielectric constant of 5.4. Repeat parts (a) and (b) in this case.

In one type of computer keyboard, each key holds a small metal plate that serves as one plate of a parallel-plate, airfilled capacitor. When the key is depressed, the plate separation decreases and the capacitance increases. Electronic circuitry detects the change in capacitance and thus detects that the key has been pressed. In one particular keyboard, the area of each metal plate is 42.0 mm\(^2\), and the separation between the plates is 0.700 mm before the key is depressed. (a) Calculate the capacitance before the key is depressed. (b) If the circuitry can detect a change in capacitance of 0.250 pF, how far must the key be depressed before the circuitry detects its depression?

A 10.0- \(\mu\)F parallel-plate capacitor with circular plates is connected to a 12.0-V battery. (a) What is the charge on each plate? (b) How much charge would be on the plates if their separation were doubled while the capacitor remained connected to the battery? (c) How much charge would be on the plates if the capacitor were connected to the 12.0-V battery after the radius of each plate was doubled without changing their separation?
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