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A \(5.80 \mu \mathrm{F}\) 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 \( 2.832 \times 10^{-2} \, \mathrm{J/m}^3 \).

Step by step solution

01

Understanding Energy Density

The energy density (u) in an electric field is given by the formula \( u = \frac{1}{2} \varepsilon E^2 \), where \( \varepsilon \) is the permittivity of the medium (for air, \( \varepsilon_0 = 8.85 \times 10^{-12} \text{ C}^2/\text{N}\cdot\text{m}^2 \)) and \( E \) is the electric field.
02

Calculating the Electric Field

The electric field \( E \) between the plates of a capacitor is given by \( E = \frac{V}{d} \), where \( V \) is the potential difference and \( d \) is the separation between the plates. Here, \( V = 400 \, \text{V} \) and \( d = 5.00 \, \text{mm} = 5.00 \times 10^{-3} \, \text{m} \). Thus, \( E = \frac{400}{5.00 \times 10^{-3}} = 8.00 \times 10^4 \, \text{V/m} \).
03

Plugging into the Energy Density Formula

Substitute \( E \) and \( \varepsilon_0 \) into the energy density formula: \( u = \frac{1}{2} \varepsilon_0 E^2 \). The calculation yields: \( u = \frac{1}{2} (8.85 \times 10^{-12}) (8.00 \times 10^4)^2 \).
04

Calculating the Result for Energy Density

Calculate the value: \( u = \frac{1}{2} (8.85 \times 10^{-12}) (6.40 \times 10^9) \), which gives \( u = 28.32 \times 10^{-3} \, \text{J/m}^3 \) or \( u = 2.832 \times 10^{-2} \, \text{J/m}^3 \).

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Key Concepts

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

Capacitance
Capacitance is a property of a capacitor that signifies its ability to store an electric charge. The capacitance is defined as the amount of charge a capacitor can store per unit voltage, represented by the formula \( C = \frac{Q}{V} \), where \( C \) is the capacitance, \( Q \) is the charge, and \( V \) is the voltage across the capacitor. For a parallel-plate capacitor, capacitance is further represented as \( C = \varepsilon \frac{A}{d} \), where \( A \) is the area of one of the plates, and \( d \) is the separation between the plates.
  • Capacitance is measured in farads (F).
  • A higher capacitance means a capacitor can store more charge at a particular voltage.
  • Capacitance depends on plate size and distance; larger areas and smaller distances increase the capacitance.
Understanding capacitance is crucial for designing circuits that require precise energy storage and distribution, especially in devices like radios and flash cameras.
Electric Field
An electric field is a region around a charged object where other charges experience a force. The strength and direction of this force are represented by the electric field \( E \). In a parallel-plate capacitor, the electric field is uniform and defined by the potential difference \( V \) and the plate separation \( d \) through the equation \( E = \frac{V}{d} \).
  • The electric field is measured in volts per meter (V/m).
  • It defines how potential energy is distributed between the charges.
  • A high electric field implies a strong force exerted on charged objects.
In a parallel-plate capacitor, the electric field is directly responsible for the potential energy stored between the plates. This energy is instrumental in circuits, influencing the performance of electronic components by controlling the movement and potential energy of charges.
Parallel-Plate Capacitor
A parallel-plate capacitor consists of two parallel metal plates separated by a dielectric (an insulating material like air). It is used to store charge via an electric field that develops between the plates when a voltage is applied. The capacity of a parallel-plate capacitor to hold charge is influenced by the area of the plates, the separation between them, and the permittivity of the dielectric.
  • Commonly used in electronic devices for energy storage, filtering, and tuning circuits.
  • Offers a practical way to manage energy density and component size.
  • Plays a critical role in regulating voltage in circuits, acting as a temporary battery.
The parallel-plate setup is fundamental in understanding basic electrical principles. Its practical applications range from simple circuit components to sophisticated energy management systems, making it an essential component in both educational and professional electronics.

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Most popular questions from this chapter

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) \(U=\frac{1}{2} C V^{2}\).

Two stationary point charges of \(+3.00 \mathrm{nC}\) and \(+2.00 \mathrm{nC}\) are separated by a distance of \(50.0 \mathrm{~cm} .\) An electron is released from rest at a point midway between the charges and moves along the line connecting them. What is the electric potential energy for the electron when it is (a) at the midpoint and (b) \(10.0 \mathrm{~cm}\) from the \(+3.00 \mathrm{nC}\) charge?

A disk-shaped parallel-plate capacitor has a capacitance \(C\). In terms of \(C,\) what would the capacitance be if the radius of each plate were increased by a factor of \(3 ?\)

\(\mathrm{A} 12.5 \mu \mathrm{F}\) capacitor is connected to a power supply that keeps a constant potential difference of \(24.0 \mathrm{~V}\) across the plates. A piece of material having a dielectric constant of 3.75 is placed between the plates, completely filling the space between them. (a) How much energy is stored in the capacitor before and after the dielectric is inserted? (b) By how much did the energy change during the insertion? Did it increase or decrease?

A \(4.00 \mu \mathrm{F}\) and a \(6.00 \mu \mathrm{F}\) capacitor are wired in parallel, and this combination is connected across a \(48.0 \mathrm{~V}\) potential difference. Calculate (a) the charge on each capacitor and (b) the potential difference across each of them.

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