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Chapter 24: Problem 25

24.25. A \(5.80-\mu F\) , parallel-plate, air capacitor has a plate separation of 5.00 \(\mathrm{mm}\) and is charged to a potential difference of 400 \(\mathrm{V}\) . Calculate the energy density in the region between the plates, in units of \(\mathrm{J} / \mathrm{m}^{3}\) .

Short Answer

Expert verified
The energy density is approximately 28.32 J/m鲁.

Step by step solution

01

Understand the Problem

We are given a parallel-plate air capacitor with a capacitance of \(5.80\,\mu F\), plate separation of \(5.00\,\text{mm}\), and a potential difference of \(400\,\text{V}\). We need to find the energy density in the region between the plates.
02

Recall the Formula for Energy Density

The energy density \( u \) of an electric field in a capacitor is given by the formula \( u = \frac{1}{2}\varepsilon E^2 \), where \( \varepsilon \) is the permittivity of the medium (air, in this case) and \( E \) is the electric field. The permittivity of free space (air) \( \varepsilon_0 \approx 8.85 \times 10^{-12} \,F/m \).
03

Calculate the Electric Field

The electric field \( E \) between the plates of a capacitor is given by \( E = \frac{V}{d} \), where \( V = 400 \, \text{V} \) is the potential difference and \( d = 5.00 \, \text{mm} = 0.005 \, \text{m} \) is the separation distance. So, \( E = \frac{400}{0.005} = 80000 \, \text{V/m} \).
04

Compute the Energy Density

Now that we have \( E = 80000 \, \text{V/m} \) and \( \varepsilon_0 = 8.85 \times 10^{-12} \, \text{F/m} \), we can compute the energy density using \( u = \frac{1}{2} \varepsilon_0 E^2 \). Substitute \( E = 80000 \, \text{V/m} \):\[ u = \frac{1}{2} \times 8.85 \times 10^{-12} \times (80000)^2 = \frac{1}{2} \times 8.85 \times 10^{-12} \times 6400000000 \approx 28.32 \, \text{J/m}^3.\]

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Key Concepts

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

Parallel-Plate Capacitor
A parallel-plate capacitor is a simple yet fundamental component in the world of electronics and physics. It consists of two large, flat, conductive plates separated by a small distance and often filled with a dielectric material.
The space between the plates stores electrical energy when a voltage is applied across them. Here's what makes them important:
  • **Storage of charge**: When a potential difference is applied to the plates, one plate becomes positively charged and the other negatively charged. This setup enables the capacitor to store electrical energy.

  • **Simple structure**: The basic structure of two parallel plates makes them easy to manufacture and model mathematically.

  • **Versatility**: Used in various electronic devices, from simple electronic circuits to complex systems.

To maximize their efficiency, the plates usually have a large surface area, minimizing the distance between them and sometimes using a dielectric to increase capacitance.
Electric Field
The electric field is a vector field that represents the force exerted by an electric charge on other charges around it. In a parallel-plate capacitor, the electric field is uniform and directed from the positive plate to the negative plate.
Understanding the electric field is crucial for determining how much energy a capacitor can store, as it directly influences the energy density.Some notable characteristics of the electric field in a parallel-plate capacitor include:
  • **Uniformity**: Inside the capacitor, the electric field is uniform, meaning it has the same magnitude and direction at every point between the plates. This uniformity simplifies calculations and theory around capacitors.

  • **Dependence on voltage and distance**: The strength of the electric field, represented with the formula \( E = \frac{V}{d} \), depends on the voltage \( V \), across the plates and their separation \( d \). As the voltage increases, the electric field strength also increases, enhancing the storage capacity.
Without a strong understanding of the electric field, it would be difficult to apply the concept of energy density.
Capacitance
Capacitance is a measure of how much charge a capacitor can store per unit voltage between its plates, represented by the formula \( C = \frac{Q}{V} \), where \( C \) is the capacitance, \( Q \) is the charge, and \( V \) is the voltage.
In practical terms, it tells us how effective the capacitor is at holding charge.The important factors affecting capacitance in a parallel-plate capacitor include:
  • **Plate area**: Larger plates can store more charge, resulting in higher capacitance.

  • **Plate separation**: Smaller distance between plates increases capacitance since it strengthens the electric field.

  • **Dielectric constant**: Materials between the plates, known as dielectrics, increase capacitance by reducing the effective electric field strength, measured by the dielectric constant.

Understanding capacitance involves recognizing how different physical factors influence a capacitor's ability to store energy, which plays a vital role in many electrical and electronic applications.

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Most popular questions from this chapter

24.2. The plates of a parallel-plate capacitor are 3.28 \(\mathrm{mm}\) apart, and each has an area of \(12.2 \mathrm{cm}^{2} .\) Each plate carries a charge of magnitude \(4.35 \times 10^{-8} \mathrm{C}\) . The plates are in vacuum. (a) What is the capacitance? (b) What is the potential difference between the plates? (c) What is the magnitude of the electric field between the plates?

24.13. A spherical capacitor contains a charge of 3.30 \(\mathrm{nC}\) when connected to a potential difference of 220 \(\mathrm{V}\) . If its plates are sepa- rated by vacuum and the inner radius of the outer shell is \(4.00 \mathrm{cm},\) calculate: (a) the capacitance; (b) the radius of the inner sphere; (c) the electric field just outside the surface of the inner sphere.

24.58. Several \(0.25-\mu \mathrm{F}\) capacitors are available. The voltage across each is not to exceed 600 \(\mathrm{V}\) . You need to make a capacitor with capacitance 0.25\(\mu \mathrm{F}\) to be connected across a potential difference of 960 \(\mathrm{V}\) . (a) Show in a diagram how an cquivalent capacitor with the desired properties can be obtained. (b) No dielectric is a perfect insulator that would not permit the flow of any charge through its volume. Suppose that the dielectric in one of the capacitors in your diagram is a moderately good conductor. What will happen in this case when your combination of capacitors is connected across the \(960-\mathbf{V}\) potential difference?

24.26. An air capacitor is made from two flat parallel plates 1.50 \(\mathrm{mm}\) apart. The magnitude of charge on each plate is 0.0180\(\mu \mathrm{C}\) when the potential difference is 200 \(\mathrm{V}\) . (a) What is the capacitance? (b) What is the area of each plate? (c) What maximum voltage can be applied without dielectric breakdown? (Dielectric breakdown for air occurs at an electric-field strength of \(3.0 \times 10^{6} \mathrm{V} / \mathrm{m}\) . (d) When the charge is 0.0180\(\mu \mathrm{C}\) , what total energy is stored?

24.53. 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 \(\frac{1}{6 / 5}\) s with an average light power output of \(2.70 \times 10^{5} \mathrm{W}\) . (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 \(\mathrm{V}\) when the stored energy equals the value calculated in part (a). What is the capacitance?
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