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Polystyrene has dielectric constant 2.6 and dielectric strength 2.0 × 107 V/m. A piece of polystyrene is used as a dielectric in a parallel-plate capacitor, filling the volume between the plates. (a) When the electric field between the plates is 80% of the dielectric strength, what is the energy density of the stored energy? (b) When the capacitor is connected to a battery with voltage 500.0 V, the electric field between the plates is 80% of the dielectric strength. What is the area of each plate if the capacitor stores 0.200 mJ of energy under these conditions?

Short Answer

Expert verified
(a) Energy density is approximately 2.95 J/m³. (b) The area of each plate is approximately 0.0032 m².

Step by step solution

01

Compute Electric Field

The electric field in the capacitor is given as 80% of the dielectric strength of polystyrene. Calculate this electric field, E, as E=0.8×2.0×107V/m=1.6×107V/m.
02

Energy Density Formula

The energy density u of the stored energy in a dielectric material can be calculated using the formula: u=12εE2 where ε=ε0×dielectric constant of polystyrene. ε0 is the permittivity of free space, 8.85×1012F/m. Substituting the values we have ε=8.85×1012×2.6.
03

Calculate Energy Density

Substituting ε and E into the energy density formula: u=12×(8.85×1012×2.6)×(1.6×107)2. Compute u to find the energy density.
04

Relate Energy to Capacitor Specifications

We know the formula for energy stored in a capacitor: U=12CV2=0.200mJ=0.200×103J. Since V=500V, we substitute U and V to find C, the capacitance.
05

Capacitance Calculation

Rearrange the energy formula to solve for C: C=U12V2=0.200×1030.5×(500)2. Calculate C.
06

Relate Capacitance to Physical Dimensions

Capacitance for a parallel-plate capacitor is given by C=εAd where d is the plate separation (can calculate from V=Ed), and A is the area of the plates. Substitute C from step 5 and solve for A.
07

Calculate Plate Area

Using d=5001.6×107 and C=0.0000016F, substitute these into the equation for capacitance C=(8.85×1012×2.6)Ad. Solve to find A, the area of each plate.

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Key Concepts

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

Dielectric Constant
The dielectric constant, also known as relative permittivity, is a measure of how much a dielectric material can increase the capacitance of a capacitor compared to when there is a vacuum between the plates. For example, the dielectric constant of polystyrene is 2.6. This means that when polystyrene is used as the dielectric material in a capacitor, the capacitance becomes 2.6 times higher than it would be without any dielectric material.

Dielectric materials are insulators, and they do not conduct electricity but can support electrostatic fields. This ability enhances the capacitor's potential to store charge, an essential aspect of various electrical applications.
  • The dielectric constant is a dimensionless number and has no units.
  • It varies depending on the material type and influences the energy-storing capacity of a capacitor.
Understanding the dielectric constant of different materials is important when designing capacitors for particular applications, as it influences both the effectiveness and efficiency of the capacitor.
Dielectric Strength
Dielectric strength is a critical property of dielectric materials, indicating the maximum electric field the material can handle before breaking down and conducting electricity. It is measured in volts per meter (V/m). In the case of polystyrene, the dielectric strength is 2.0 × 10⁷ V/m.

When a dielectric material reaches or exceeds its dielectric strength, it can no longer function as an insulator and may allow the enclosed charges to move freely across the material, causing a short circuit or damage to the capacitor.
  • Dielectric strength ensures that under normal operating conditions, the material prevents breakdown and maintains its insulating properties.
  • It is vital to select a material with a sufficient dielectric strength for applications with high voltages to prevent failure.
When designing or using capacitors, always ensure that the operational electric field stays within a safe margin below the dielectric strength to ensure longevity and reliability.
Energy Density
Energy density is the amount of energy stored per unit volume within a capacitor. It is an important measure indicating how effectively a capacitor can store energy in the field it creates. In formula terms, the energy density u is given by u=12εE2where ε is the permittivity of the dielectric material, and E is the electric field strength.

For polystyrene with a dielectric constant of 2.6 and an electric field of 1.6 × 10⁷ V/m, we substitute these values to calculate the energy density. This expresses how the internal structure of a capacitor is capable of storing energy given the specific properties of the material used.
  • Higher permittivity and dielectric constants allow for greater energy density.
  • Energy density is a critical factor in applications where size and efficiency are constraints.
Understanding energy density helps in designing capacitors that are both compact and efficient, catering to advanced technology solutions.
Parallel-Plate Capacitor
The parallel-plate capacitor is one of the most simple and commonly used capacitor designs. It consists of two conductive plates separated by a dielectric material. The key feature of this capacitor is that the capacitance C is directly related to the area of the plates A, the separation distance d, and the dielectric constant εr of the material, with the formula:
C=ε0εrAd
where ε0 is the permittivity of free space.

The parallel-plate capacitor serves a wide array of applications due to its straightforward design and reliable functionality. Typically, it is used whenever a precise and stable capacitance value is needed, as the physical parameters influencing the capacitance can be accurately controlled.
  • The area of the plates and the nature of the dielectric material directly affect the overall capacitance.
  • The simplicity of the design makes it easy to manufacture and integrate into various circuits.
Designing with parallel-plate capacitors requires understanding how changes to the physical structure or materials impact the overall behaviour and efficiency of the capacitor.

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Most popular questions from this chapter

Electric eels and electric fish generate large potential differences that are used to stun enemies and prey. These potentials are produced by cells that each can generate 0.10 V. We can plausibly model such cells as charged capacitors. (a) How should these cells be connected (in series or in parallel) to produce a total potential of more than 0.10 V? (b) Using the connection in part (a), how many cells must be connected together to produce the 500-V surge of the electric eel?

When a 360-nF air capacitor (11 nF = 109 F) is connected to a power supply, the energy stored in the capacitor is 1.85 × 105 J. While the capacitor is kept connected to the power supply, a slab of dielectric is inserted that completely fills the space between the plates. This increases the stored energy by 2.32 × 105 J. (a) What is the potential difference between the capacitor plates? (b) What is the dielectric constant of the slab?

A capacitor is made from two hollow, coaxial, iron cylinders, one inside the other. The inner cylinder is negatively charged and the outer is positively charged; the magnitude of the charge on each is 10.0 pC. The inner cylinder has radius 0.50 mm, the outer one has radius 5.00 mm, and the length of each cylinder is 18.0 cm. (a) What is the capacitance? (b) What applied potential difference is necessary to produce these charges on the cylinders?

A 5.00μF 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?

A 0.350-m-long cylindrical capacitor consists of a solid conducting core with a radius of 1.20 mm and an outer hollow conducting tube with an inner radius of 2.00 mm. The two conductors are separated by air and charged to a potential difference of 6.00 V. Calculate (a) the charge per length for the capacitor; (b) the total charge on the capacitor; (c) the capacitance; (d) the energy stored in the capacitor when fully charged.

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