Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

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.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

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.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Your electronics company has several identical capacitors with capacitance C1 and several others with capacitance C2. You must determine the values of C1 and C2 but don't have access to C1 and C2 individually. Instead, you have a network with C1 and C2 connected in series and a network with C1 and C2 connected in parallel. You have a 200.0-V battery and instrumentation that measures the total energy supplied by the battery when it is connected to the network. When the parallel combination is connected to the battery, 0.180 J of energy is stored in the network. When the series combination is connected, 0.0400 J of energy is stored. You are told that C1 is greater than C2. (a) Calculate C1 and C2. (b) For the series combination, does C1 or C2 store more charge, or are the values equal? Does C1 or C2 store more energy, or are the values equal? (c) Repeat part (b) for the parallel combination.

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× 106 V/m. What is (a) the potential difference between the plates; (b) the area of each plate; (c) the capacitance?

Suppose that the change in Vm was caused by the entry of Ca2+ instead of Na+. How many Ca2+ ions would have to enter the cell per unit membrane to produce the change? (a) Half as many as for Na+; (b) the same as for Na+; (c) twice as many as for Na+; (d) cannot say without knowing the inside and outside concentrations of Ca2+.

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?

Two parallel plates have equal and opposite charges. When the space between the plates is evacuated, the electric field is E=3.20×105 V/m. When the space is filled with dielectric, the electric field is E=2.50×105 V/m. (a) What is the charge density on each surface of the dielectric? (b) What is the dielectric constant?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free