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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

Electronic flash units for cameras contain a capacitor for storing the energy used to produce the flash. In one such unit, the flash lasts for \({1 \over 675}\) s with an average light power output of 2.70 \(\times\) 10\(^5\) W. (a) If the conversion of electrical energy to light is 95% efficient (the rest of the energy goes to thermal energy), how much energy must be stored in the capacitor for one flash? (b) The capacitor has a potential difference between its plates of 125 V when the stored energy equals the value calculated in part (a). What is the capacitance?

A 5.00\(\mu\)F parallel-plate capacitor is connected to a 12.0-V battery. After the capacitor is fully charged, the battery is disconnected without loss of any of the charge on the plates. (a) A voltmeter is connected across the two plates without discharging them. What does it read? (b) What would the voltmeter read if (i) the plate separation were doubled; (ii) the radius of each plate were doubled but their separation was unchanged?

The plates of a parallel-plate capacitor are 2.50 mm apart, and each carries a charge of magnitude 80.0 nC. The plates are in vacuum. The electric field between the plates has a magnitude of 4.00\(\times\) 10\(^6\) V/m. What is (a) the potential difference between the plates; (b) the area of each plate; (c) the capacitance?

A parallel-plate vacuum capacitor has 8.38 J of energy stored in it. The separation between the plates is 2.30 mm. If the separation is decreased to 1.15 mm, what is the energy stored (a) if the capacitor is disconnected from the potential source so the charge on the plates remains constant, and (b) if the capacitor remains connected to the potential source so the potential difference between the plates remains constant?

Two parallel plates have equal and opposite charges. When the space between the plates is evacuated, the electric field is \(E = 3.20\times10^5\space V/m\). When the space is filled with dielectric, the electric field is \(E = 2.50\times10^5\space V/m\). (a) What is the charge density on each surface of the dielectric? (b) What is the dielectric constant?

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