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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 V. Calculate the energy density in the region between the plates, in units of J/m\(^3\).

Short Answer

Expert verified
The energy density is approximately 0.0283 J/m³.

Step by step solution

01

Identify the Formula for Energy Density

The energy density (u) in the region between the plates of a capacitor is given by the formula: \[ u = \frac{1}{2} \varepsilon_0 E^2 \] where \( \varepsilon_0 \) is the permittivity of free space (\( 8.85 \times 10^{-12} \ \text{C}^2/(\text{N}\cdot\text{m}^2) \)) and \( E \) is the electric field strength between the plates.
02

Calculate the Electric Field

The electric field \( E \) between parallel plates is given by the formula: \[ E = \frac{V}{d} \] where \( V \) is the potential difference across the plates (400 V) and \( d \) is the separation between the plates (5.00 mm or 0.005 m). Substitute these values into the formula: \[ E = \frac{400 \ \text{V}}{0.005 \ \text{m}} = 80,000 \ \text{V/m} \].
03

Substitute into the Energy Density Formula

Substitute the value of \( E \) and \( \varepsilon_0 \) into the energy density formula: \[ u = \frac{1}{2} \times 8.85 \times 10^{-12} \ \text{C}^2/(\text{N}\cdot\text{m}^2) \times (80,000 \ \text{V/m})^2 \].
04

Compute the Energy Density

Calculate the numerical value of the energy density: \[ u = \frac{1}{2} \times 8.85 \times 10^{-12} \times 6.4 \times 10^{9} = 2.83 \times 10^{-2} \ \text{J/m}^3 \].

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Key Concepts

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

Parallel-Plate Capacitor
A parallel-plate capacitor is a simple device that consists of two flat, conductive plates separated by a distance.
It stores electrical energy by maintaining a voltage across the plates. This arrangement sets an electric field in the space between them, creating a storage field for energy.
Parallel-plate capacitors are often used in electronics because of their straightforward design and ability to store an electric charge efficiently.
When a voltage is applied, the capacitor charges up with positive charges on one plate and negative charges on the other, with an electric field established between them.
  • The stored energy in a capacitor is crucial for controlling power supply and processing electronic signals in circuits.
  • The capacitance, which indicates how much charge it can store, is influenced by the plate area, separation distance, and the medium between the plates.
Understanding these basics aids in grasping how capacitors function within larger electrical systems.
Electric Field Calculation
The electric field in a parallel-plate capacitor is uniform and calculated using the voltage across the plates and the distance separating them.
This relation is given by the formula: \[ E = \frac{V}{d} \]
where \( E \) is the electric field strength, \( V \) is the potential difference, and \( d \) is the plate separation.
In the context of the exercise, for a voltage of 400 V and a plate separation of 0.005 m (converted from 5.00 mm), the electric field can be calculated by substituting the values:
\[ E = \frac{400 \ \text{V}}{0.005 \ \text{m}} = 80,000 \ \text{V/m}\].
  • Knowing the electric field helps understand the force exerted on charges between the plates.
  • It's a key factor in determining the energy density—a measure of energy stored per unit volume in the capacitor.
This calculation highlights the strong influence of plate separation on the electric field and energy storage.
Permittivity of Free Space
The permittivity of free space, also known as \( \varepsilon_0 \), is a fundamental physical constant that describes how electric fields interact with a vacuum.
Its value is approximately \( 8.85 \times 10^{-12} \ \text{C}^2/(\text{N}\cdot\text{m}^2)\). This constant is crucial when analyzing capacitors and their ability to store energy.
In the formula for the energy density of a parallel-plate capacitor, \( u = \frac{1}{2} \varepsilon_0 E^2 \), \( \varepsilon_0 \) acts as a scaling factor, affecting how much energy is stashed between the plates for a given electric field.
The permittivity of free space defines the influence of a vacuum environment on the electric field strength and energy storage efficiency.
  • Higher permittivity means greater capacitance and energy storage capability in a capacitor.
  • It also plays an essential role in determining the propagation of electromagnetic waves.
Understanding \( \varepsilon_0 \) is central to grasping the behavior of electric and magnetic fields in various environments.

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Most popular questions from this chapter

A parallel-plate capacitor with only air between the plates is charged by connecting it to a battery. The capacitor is then disconnected from the battery, without any of the charge leaving the plates. (a) A voltmeter reads 45.0 V when placed across the capacitor. When a dielectric is inserted between the plates, completely filling the space, the voltmeter reads 11.5 V. What is the dielectric constant of this material? (b) What will the voltmeter read if the dielectric is now pulled partway out so it fills only one-third of the space between the plates?

The dielectric to be used in a parallel-plate capacitor has a dielectric constant of 3.60 and a dielectric strength of 1.60 \(\times\) 10\(^7\) V>m. The capacitor is to have a capacitance of 1.25 \(\times\) 10\(^{-9}\) F and must be able to withstand a maximum potential difference of 5500 V. What is the minimum area the plates of the capacitor may have?

A parallel-plate capacitor has capacitance \(C_0\) = 8.00 pF when there is air between the plates. The separation between the plates is 1.50 mm. (a) What is the maximum magnitude of charge Q that can be placed on each plate if the electric field in the region between the plates is not to exceed 3.00 \(\times\) 10\(^4\) V/m? (b) A dielectric with \(K = 2.70\) is inserted between the plates of the capacitor, completely filling the volume between the plates. Now what is the maximum magnitude of charge on each plate if the electric field between the plates is not to exceed 3.00 \(\times\) 10\(^4\) V/m?

A 12.5-\(\mu\)F capacitor is connected to a power supply that keeps a constant potential difference of 24.0 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 capacitor is formed from two concentric spherical conducting shells separated by vacuum. The inner sphere has radius 12.5 cm, and the outer sphere has radius 14.8 cm. A potential difference of 120 V is applied to the capacitor. (a) What is the energy density at \(r\) = 12.6 cm, just outside the inner sphere? (b) What is the energy density at \(r\) = 14.7 cm, just inside the outer sphere? (c) For a parallel-plate capacitor the energy density is uniform in the region between the plates, except near the edges of the plates. Is this also true for a spherical capacitor?

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