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A 5.80-μF, 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/m3.

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=12ε0E2 where ε0 is the permittivity of free space (8.85×1012 C2/(Nm2)) 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=Vd 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=400 V0.005 m=80,000 V/m.
03

Substitute into the Energy Density Formula

Substitute the value of E and ε0 into the energy density formula: u=12×8.85×1012 C2/(Nm2)×(80,000 V/m)2.
04

Compute the Energy Density

Calculate the numerical value of the energy density: u=12×8.85×1012×6.4×109=2.83×102 J/m3.

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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=Vd
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=400 V0.005 m=80,000 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 ε0, is a fundamental physical constant that describes how electric fields interact with a vacuum.
Its value is approximately 8.85×1012 C2/(Nm2). 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=12ε0E2, ε0 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 ε0 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 cylindrical capacitor consists of a solid inner conducting core with radius 0.250 cm, surrounded by an outer hollow conducting tube. The two conductors are separated by air, and the length of the cylinder is 12.0 cm. The capacitance is 36.7 pF. (a) Calculate the inner radius of the hollow tube. (b) When the capacitor is charged to 125 V, what is the charge per unitlength λ on the capacitor?

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 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 air capacitor of capacitance 245 pF has a charge of magnitude 0.148 μC on each plate. The plates are 0.328 mm apart. (a) What is the potential difference between the plates? (b) What is the area of each plate? (c) What is the electricfield magnitude between the plates? (d) What is the surface charge density on each plate?

A parallel-plate capacitor has capacitance C0 = 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 × 104 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 × 104 V/m?

A spherical capacitor is formed from two concentric, spherical, conducting shells separated by vacuum. The inner sphere has radius 15.0 cm and the capacitance is 116 pF. (a) What is the radius of the outer sphere? (b) If the potential difference between the two spheres is 220 V, what is the magnitude of charge on each sphere?

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