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Chapter 26: Problem 36

A uniform electric field \(E=3000 \mathrm{V} / \mathrm{m}\) exists within a certain region. What volume of space contains an energy equal to \(1.00 \times 10^{-7} \mathrm{J}\) ? Express your answer in cubic meters and in liters.

Short Answer

Expert verified
The volume of space is approximately \( 3.37 \times 10^{-6} \,\text{m}^3 \) or \( 3.37 \) liters.

Step by step solution

01

Recall the energy density of an electric field

The energy density of an electric field (energy per unit volume) is given by the formula \( u = \frac{1}{2} \epsilon_0 E^2 \), where \( \epsilon_0 \) is the vacuum permittivity constant with a value of \( 8.85 \times 10^{-12} \,\text{C}^2/\text{N}\cdot\text{m}^2 \), and \( E \) is the electric field strength.
02

Calculate the energy density

Substitute the given value of the electric field \( E = 3000 \,\text{V/m} \) into the energy density formula to calculate the energy per unit volume.
03

Determine the volume that contains the given energy

To find the volume that contains an energy of \( 1.00 \times 10^{-7} \,\text{J} \), divide the total energy by the energy density obtained from Step 2.
04

Convert the volume to liters

Since 1 liter equals to \( 1 \times 10^{-3} \) cubic meters, convert the volume from cubic meters to liters by multiplying it with \( 1000 \).

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Key Concepts

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

Uniform Electric Field
Imagine a space where all points experience the same magnitude and direction of force from an electric field; this is what we refer to as a uniform electric field. In such a field, if you were to place a positive test charge, it would move in a straight line, accelerated by a constant force. For students tackling problems in electromagnetism, picturing this uniformity can simplify the situation significantly. Unlike variable fields, which require complex calculus to solve, uniform fields' properties can be understood and applied using much simpler algebra.
Vacuum Permittivity
The term vacuum permittivity, represented by the symbol \( \epsilon_0 \), is a constant fundamental to electromagnetism. It describes how easily an electric field can permeate through a vacuum. The value of \( \epsilon_0 \) is approximately \( 8.85 \times 10^{-12} \,\text{C}^2/\text{N}\cdot\text{m}^2 \). This constant plays a crucial role in calculating the force between charged particles and the energy stored within an electric field in a vacuum. It is the bedrock upon which many equations in electromagnetism rest, serving as a scaling factor that adjusts the theoretical models to match real-world observations.
Energy Per Unit Volume
The idea of energy per unit volume in the context of an electric field conveys how much energy is stored in every cubic meter of the field. This concept is critical when considering the energy stored in capacitors or the energy distribution in electrical systems. The energy density \( u \) is calculated using the formula \[ u = \frac{1}{2} \epsilon_0 E^2 \], which associates the energy density directly with the strength of the electric field squared. This relationship implies that even small increases in electric field strength result in a quadratically higher energy density, showcasing how sensitive energy storage is to the field's intensity.
Electric Field Strength
When we discuss electric field strength, symbolized as \( E \), we are referring to the force per unit charge that a small positive test charge would experience within the field. It's measured in volts per meter (V/m) in the International System of Units (SI). A strong electric field signifies a substantial force acting on charges within that region; conversely, a weak field applies less force. The strength of the electric field not only influences the motion of charges but also relates to how much energy the field can store, as reflected in the energy density formula.
Volume Conversion
Working with different units of volume can often be necessary when we study the physics of materials and fluids. In the original exercise, we transition from the metric unit of cubic meters to the more commonly used liters for everyday contexts. It's crucial for students to master the conversion, which is simply that 1 liter is equivalent to \( 1 \times 10^{-3} \) cubic meters. Especially in chemistry and biology, this volume conversion is a fundamental skill, as it allows one to seamlessly switch between scales, from objects' microscopic internal spaces to larger, tangible volumes of liquids.

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Most popular questions from this chapter

(a) Two spheres have radii \(a\) and \(b\) and their centers are a distance \(d\) apart. Show that the capacitance of this system is $$C=\frac{4 \pi \epsilon_{0}}{\frac{1}{a}+\frac{1}{b}-\frac{2}{d}}$$ provided that \(d\) is large compared with \(a\) and \(b\) . (Suggestion: Because the spheres are far apart, assume that the potential of each equals the sum of the potentials due to each sphere, and when calculating those potentials assume that \(V=k_{e} Q / r\) applies. (b) Show that as \(d\) approaches infinity the above result reduces to that of two spherical capacitors in series.

To repair a power supply for a stereo amplifier, an electronics technician needs a \(100-\mu \mathrm{F}\) capacitor capable of withstanding a potential difference of 90 \(\mathrm{V}\) between the plates. The only available supply is a box of five \(100-\mu \mathrm{F}\) capacitors, each having a maximum voltage capability of 50 V. Can the technician substitute a combination of these capacitors that has the proper electrical characteristics? If so, what will be the maximum voltage across any of the capacitors used? (Suggestion: The technician may not have to use all the capacitors in the box.)

One conductor of an overhead electric transmission line is a long aluminum wire 2.40 \(\mathrm{cm}\) in radius. Suppose that at a particular moment it carries charge per length 1.40\(\mu \mathrm{C} / \mathrm{m}\) and is at potential 345 \(\mathrm{kV}\) . Find the potential 12.0 \(\mathrm{m}\) below the wire. Ignore the other conductors of the transmission line and assume the electric field is everywhere purely radial.

The general form of Gauss's law describes how a charge creates an electric field in a material, as well as in vacuum. It is $$\oint \mathbf{E} \cdot d \mathbf{A}=\frac{q}{\epsilon}$$ where \(\epsilon=\kappa \epsilon_{0}\) is the permittivity of the material. (a) A sheet with charge \(Q\) uniformly distributed over its area \(A\) is surrounded by a dielectric. Show that the sheet creates a uniform electric field at nearby points, with magnitude \(E=Q / 2 A \epsilon\) . (b) Two large sheets of area \(A,\) carrying opposite charges of equal magnitude \(Q,\) are a small distance \(d\) apart. Show that they create uniform electric field in the space between them, with magnitude \(E=Q / A \epsilon\) (c) Assume that the negative plate is at zero potential. Show that the positive plate is at potential \(Q d / A \epsilon\) (d) Show that the capacitance of the pair of plates is \(A \epsilon / d=\kappa A \epsilon_{0} / d .\)

The immediate cause of many deaths is ventricular fibrillation, uncoordinated quivering of the heart as opposed to proper beating. An electric shock to the chest can cause momentary paralysis of the heart muscle, after which the heart will sometimes start organized beating again. A defibrilator (Fig. 26.14\()\) is a device that applies a strong electric shock to the chest over a time interval of a few milliseconds. The device contains a capacitor of several microfarads, charged to several thousand volts. Electrodes called paddles, about 8 \(\mathrm{cm}\) across and coated with conducting paste, are held against the chest on both sides of the heart. Their handles are insulated to prevent injury to the operator, who calls, "Clear!" and pushes a button on one paddle to discharge the capacitor through the patient's chest. Assume that an energy of 300 \(\mathrm{J}\) is to be delivered from a \(30.0-\mu \mathrm{F}\) capacitor. To what potential difference must it be charged?
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