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A \(5.80-\mu \mathrm{F}\), parallel-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\(^3\).

Step by step solution

01

Understand the Formula for Energy Density

The energy density (u) in a capacitor is given by the formula \( u = \frac{1}{2} \varepsilon E^2 \), where \( \varepsilon \) is the permittivity of the medium between the plates (for air, \( \varepsilon_0 \approx 8.85 \times 10^{-12} \mathrm{F/m} \)), and \( E \) is the electric field between the plates.
02

Calculate the Electric Field (E)

The electric field \( E \) in a parallel-plate capacitor is given by \( E = \frac{V}{d} \), where \( V = 400 \mathrm{V} \) and \( d = 5.00 \times 10^{-3} \mathrm{m} \). Therefore, \( E = \frac{400}{5.00 \times 10^{-3}} = 80,000 \mathrm{V/m} \).
03

Compute the Permittivity of Air

The permittivity of free space (or air) \( \varepsilon_0 \) is approximately \( 8.85 \times 10^{-12} \mathrm{F/m} \). This is a constant value that will be used in the calculation of energy density.
04

Calculate the Energy Density (u)

Using the formula for energy density \( u = \frac{1}{2} \varepsilon_0 E^2 \) and substituting the known values: \( u = \frac{1}{2} \times 8.85 \times 10^{-12} \times (80,000)^2 \). Calculating this gives \( u \approx 28.32 \) 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
Imagine a simple setup where two flat plates face each other with some distance between them. This is what we call a parallel-plate capacitor. It's a basic yet powerful structure used in electronics. Capacitors store energy, and their setup allows them to hold electric charge. The plates are separated by an insulating material, often air or another dielectric. This arrangement allows them to store electrical energy. Key features of parallel-plate capacitors:
  • The plates are identical in size and shape, usually circular or rectangular.
  • The space between the plates can affect the capacitor's ability to hold charge.
  • A dielectric material between the plates can increase the capacitor's capacity by reducing the effective distance between the plates.
Parallel-plate capacitors are commonly used in circuits to help manage voltage and power levels. They play a crucial role in tuning radios, filtering noises, and stabilizing power supplies.
Electric Field
An electric field is a fascinating concept in physics, controlling how electric charges interact. In a parallel-plate capacitor, the electric field is the force per unit charge that appears between the plates. Think of it like invisible lines of force pushing charges around.The strength of this field depends on:
  • The potential difference applied to the plates, which we measure in volts (V).
  • The distance between the plates, known as the separation distance.
We calculate the electric field's magnitude in a parallel-plate capacitor with the formula: \[ E = \frac{V}{d} \]where \( V \) is the potential difference and \( d \) is the separation between plates.This setup creates a uniform electric field with a consistent strength and direction between the plates. It's this electric field that holds the capacitor's potential energy, allowing it to function in electronic circuits.
Permittivity of Free Space
The permittivity of free space is a critical concept when discussing electric fields and capacitors. Often denoted as \( \varepsilon_0 \), it quantifies how much electric field can pass through a vacuum. For most practical purposes involving air, we consider its value:\( \varepsilon_0 \approx 8.85 \times 10^{-12} \text{ F/m} \)Why is this important? The permittivity value helps determine how effectively a capacitor can store energy. It is part of the formula to calculate energy density in a capacitor, helping to bridge its physical characteristics with its electrical properties.This constant represents how much resistance the electric field encounters as it disperses through the vacuum. A higher permittivity value means the material allows more electric lines of force, which means a better capacity to store energy.
Potential Difference
In essence, a potential difference, or voltage, is what drives the electric field within a capacitor. It's the force that pushes charges around, affecting how much energy the capacitor can store. Key points about potential difference:
  • We measure it in volts (V).
  • It's crucial for determining the energy stored in a capacitor.
  • The greater the potential difference, the stronger the electric field between the plates.
In a parallel-plate capacitor, the potential difference is applied across the plates, creating an electric field. This field directly influences the capacitor's ability to store energy. Higher voltage means a more considerable amount of charge can be stored, which results in a higher energy density. Understanding potential difference is vital for designing circuits and selecting the right capacitors for specific electronic applications.

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Most popular questions from this chapter

You are working on an electronics project requiring a variety of capacitors, but you have only a large supply of 100 -nF capacitors available. Show how you can connect these capacitors to produce each of the following equivalent capacitances: (a) 50 \(\mathrm{nF}\) ; (b) \(450 \mathrm{nF} ;\) (c) \(25 \mathrm{nF} ;\) (d) 75 \(\mathrm{nF}\) .

A parallel-plate capacitor has plates with area 0.0225 \(\mathrm{m}^{2}\) separated by 1.00 \(\mathrm{mm}\) of Teflon. (a) Calculate the charge on the plates when they are charged to a potential difference of 12.0 \(\mathrm{V}\) . (b) Use Gauss's law (Eq. 24.23 ) to calculate the electric field inside the Teflon. (c) Use Gauss's law to calculate the electric field if the voltage source is disconnected and the Teflon is removed.

An air capacitor is made by using two flat plates, each with area \(A,\) separated by a distance \(d\) . Then a a metal slab having thickness a (less than \(d\) ) and the same shape and size as the plates is inserted between them, parallel to the plates and not touching either plate (Fig. P24.66). (a) What is the capacitance of this arrangement? (b) Express the capacitance as a multiple of the capacitance \(C_{0}\) when the metal slab is not present. (c) Discuss what happens to the capacitance in the limits \(a \rightarrow 0\) and \(a \rightarrow d\) .

A cylindrical air capacitor of length 15.0 \(\mathrm{m}\) stores \(3.20 \times 10^{-9} \mathrm{J}\) of energy when the potential difference between the two conductors is 4.00 \(\mathrm{V}\) . (a) Calculate the magnitude of the charge on each conductor. (b) Calculate the ratio of the radii of the inner and outer conductors.

A cylindrical capacitor consists of a solid inner conducting core with radius \(0.250 \mathrm{cm},\) surrounded by an outer hollow conducting tube. The two conductors are separated by air, and the length of the cylinder is 12.0 \(\mathrm{cm}\) . The capacitance is 36.7 pF. (a) Calculate the inner radius of the hollow tube. (b) When the capacitor is charged to \(125 \mathrm{V},\) what is the charge per unit length \(\lambda\) on the capacitor?

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