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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 budding electronics hobbyist wants to make a simple 1.0-nF capacitor for tuning her crystal radio, using two sheets of aluminum foil as plates, with a few sheets of paper between them as a dielectric. The paper has a dielectric constant of 3.0, and the thickness of one sheet of it is 0.20 mm. (a) If the sheets of paper measure 22 \(\times\)28 cm and she cuts the aluminum foil to the same dimensions, how many sheets of paper should she use between her plates to get the proper capacitance? (b) Suppose for convenience she wants to use a single sheet of posterboard, with the same dielectric constant but a thickness of 12.0 mm, instead of the paper. What area of aluminum foil will she need for her plates to get her 1.0 nF of capacitance? (c) Suppose she goes high-tech and finds a sheet of Teflon of the same thickness as the posterboard to use as a dielectric. Will she need a larger or smaller area of Teflon than of posterboard? Explain.

A 10.0- \(\mu\)F parallel-plate capacitor with circular plates is connected to a 12.0-V battery. (a) What is the charge on each plate? (b) How much charge would be on the plates if their separation were doubled while the capacitor remained connected to the battery? (c) How much charge would be on the plates if the capacitor were connected to the 12.0-V battery after the radius of each plate was doubled without changing their separation?

Cell membranes (the walled enclosure around a cell) are typically about 7.5 nm thick. They are partially permeable to allow charged material to pass in and out, as needed. Equal but opposite charge densities build up on the inside and outside faces of such a membrane, and these charges prevent additional charges from passing through the cell wall. We can model a cell membrane as a parallel-plate capacitor, with the membrane itself containing proteins embedded in an organic material to give the membrane a dielectric constant of about 10. (See \(\textbf{Fig. P24.48}\).) (a) What is the capacitance per square centimeter of such a cell wall? (b) In its normal resting state, a cell has a potential difference of 85 mV across its membrane. What is the electric field inside this membrane?

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?

A cylindrical capacitor has an inner conductor of radius 2.2 mm and an outer conductor of radius 3.5 mm. The two conductors are separated by vacuum, and the entire capacitor is 2.8 m long. (a) What is the capacitance per unit length? (b) The potential of the inner conductor is 350 mV higher than that of the outer conductor. Find the charge (magnitude and sign) on both conductors.

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