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Chapter 18: Problem 58

A 5.80\(\mu\) F parallel-plate air capacitor has a plate separation of 5.00 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 \(2.832 \times 10^{-3} \text{ J/m}^3\).

Step by step solution

01

Understand the Concept

The energy density (\(u\) represents the energy stored per unit volume in an electric field. For a parallel-plate capacitor, \(u\) is given by the formula \(u = \frac{1}{2} \epsilon_0 E^2\), where \(\epsilon_0\) is the permittivity of free space and \(E\) is the electric field between the plates.
02

Calculate the Electric Field

The electric field \(E\) between the plates of a parallel-plate capacitor is given by \(E = \frac{V}{d}\), where \(V\) is the potential difference (400 V) and \(d\) is the separation between the plates (5.00 mm or 0.005 m). Calculate \(E\):\[E = \frac{400}{0.005} = 80000\, \text{V/m}\]
03

Apply the Energy Density Formula

Use the energy density formula \(u = \frac{1}{2} \epsilon_0 E^2\), where \(\epsilon_0 = 8.85 \times 10^{-12} \text{ F/m}\). Substitute the value of \(E\) calculated in the previous step:\[u = \frac{1}{2} \times 8.85 \times 10^{-12} \times (80000)^2\]Simplify this expression to find \(u\).
04

Complete Calculation

Finish the calculation:\[u = \frac{1}{2} \times 8.85 \times 10^{-12} \times 6400000000\]\[u = 2.832 \times 10^{-3} \text{ J/m}^3\]
05

Verify and Interpret

The energy density calculated is \(2.832 \times 10^{-3} \text{ J/m}^3\). This represents the amount of energy stored per unit volume between the capacitor plates when charged to the given potential difference.

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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 consists of two plates separated by a small distance. It's like a special device that can store electrical energy. You place one plate with a positive charge and the other with a negative charge.
This separation creates an electric field between them, which stores energy. Capacitors are used in many electronic devices, such as radios and TVs, to ensure a steady supply of energy.
  • The plates are usually large and positioned close together to maximize the capacity to hold charge.
  • The separation of the plates is filled with an insulator, which is often air, and this helps to prevent an electrical discharge between the plates.
  • The ability of a capacitor to store charge is measured in Farads, which depends on the size of the plates and their separation.
Understanding these properties can help you predict how a capacitor will behave in a circuit.
Electric Field
An electric field is an invisible force field that surrounds electric charges. Think of it like a wind that can push objects around, except it's for charges. When dealing with a parallel-plate capacitor, the electric field ( E)
Potential Difference
represents the force that pushes positive charges from one plate to another.
The strength of this electric field in a capacitor is calculated using the formula \(E = \frac{V}{d}\), where \(V\) is the potential difference across the plates, and \(d\) is the separation distance between them.
This electric field plays a crucial role in calculating energy density, as it directly influences the amount of energy stored in the capacitor.
  • It determines how much work is needed to move a charge between the plates.
  • The stronger the electric field, the more energy the capacitor can store.
  • Electric fields can influence how a device or circuit operates, especially when they change rapidly.
Becoming familiar with electric fields helps in understanding the behavior of charges in various systems.
Permittivity of Free Space
Permittivity of free space, denoted as \(\epsilon_0\), is a constant that describes how well the space can "permit" electric field lines. Picture space occupied by an electric field without any interference.
It's a fundamental constant in physics, with a value of \(8.85 \times 10^{-12} \text{ F/m}\). This value is important when calculating how capacitors store energy.
This factor impacts how capacitors operate in electric circuits, influencing their effectiveness and efficiency.
  • \(\epsilon_0\) plays a role in the energy density formula, \(u = \frac{1}{2} \epsilon_0 E^2\).
  • The lower the permittivity, the less electric field lines will interact with the medium.
  • Understanding permittivity helps in designing capacitors that perform well in circuits.
This essential constant is key in the formulas that predict how a capacitor will perform, ensuring you can make accurate calculations in physics problems.

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Most popular questions from this chapter

Cathode-ray tube. A cathode-ray tube (CRT) is an evacuated glass tube. Electrons are produced at one end, usually by the heating of a metal. After being focused electromagnetically into a beam, they are accelerated through a potential difference, called the accelerating potential. The electrons then strike a coated screen, where they transfer their energy to the coating through collisions, causing it to glow. CRTs are found in oscilloscopes and computer monitors, as well as in earlier versions of television screens. (a) If an electron of mass \(m\) and charge \(-e\) is accelerated from rest through an accelerating potential \(V,\) show that the speed it gains is \(v=\sqrt{2 e V / m}\) . We are assuming that \(V\) is small enough that the final speed is much less than the speed of light.) (b) If the accelerating potential is \(95 \mathrm{V},\) how fast will the electrons be moving when they hit the screen?

Electrical sensitivity of sharks. Certain sharks can detect an electric field as weak as 1.0\(\mu \mathrm{V} / \mathrm{m} .\) To grasp how weak this field is, if you wanted to produce it between two parallel metal plates by connecting an ordinary 1.5 \(\mathrm{V}\) A battery across theseplates, how far apart would the plates have to be?

In the text, it was shown that the energy stored in a capacitor \(C\) charged to a potential \(V\) is \(U=\frac{1}{2} Q V\) . Show that this energy can also be expressed as (a) \(U=Q^{2} / 2 C\) and (b) \(U=\frac{1}{2} C V^{2}\)

A \(\mathrm{A} 20.0 \mu \mathrm{F}\) capacitor is charged to a potential difference of 800 \(\mathrm{V} .\) The terminals of the charged capacitor are then connected to those of an uncharged 10.0\(\mu \mathrm{F}\) capacitor. Compute (a) the onginal charge of the system, (b) the final potential difference across each capacitor, (c) the final energy of the system, and (d) the decrease in energy when the capacitors are connected.

At a certain distance from a point charge, the potential and electric-field magnitude due to that charge are 4.98 V and \(12.0 \mathrm{V} / \mathrm{m},\) respectively. (Take the potential to be zero at infinity.) (a) What is the distance to the point charge? (b) What is the magnitude of the charge? (c) Is the electric field directed toward or away from the point charge?
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