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Polystyrene has dielectric constant 2.6 and dielectric strength 2.0 \(\times\) 10\(^7\) 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 \( E = 0.8 \times 2.0 \times 10^7 \, \text{V/m} = 1.6 \times 10^7 \, \text{V/m} \).
02

Energy Density Formula

The energy density \( u \) of the stored energy in a dielectric material can be calculated using the formula: \[ u = \frac{1}{2} \varepsilon E^2 \] where \( \varepsilon = \varepsilon_0 \times \text{dielectric constant of polystyrene} \). \( \varepsilon_0 \) is the permittivity of free space, \( 8.85 \times 10^{-12} \, \text{F/m} \). Substituting the values we have \( \varepsilon = 8.85 \times 10^{-12} \times 2.6 \).
03

Calculate Energy Density

Substituting \( \varepsilon \) and \( E \) into the energy density formula: \[ u = \frac{1}{2} \times (8.85 \times 10^{-12} \times 2.6) \times (1.6 \times 10^7)^2 \]. Compute \( u \) to find the energy density.
04

Relate Energy to Capacitor Specifications

We know the formula for energy stored in a capacitor: \[ U = \frac{1}{2} CV^2 = 0.200 \, \text{mJ} = 0.200 \times 10^{-3} \, \text{J} \]. Since \( V = 500 \, \text{V} \), we substitute \( U \) and \( V \) to find \( C \), the capacitance.
05

Capacitance Calculation

Rearrange the energy formula to solve for \( C \): \[ C = \frac{U}{\frac{1}{2} V^2} = \frac{0.200 \times 10^{-3}}{0.5 \times (500)^2} \]. Calculate \( C \).
06

Relate Capacitance to Physical Dimensions

Capacitance for a parallel-plate capacitor is given by \( C = \frac{\varepsilon A}{d} \) 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 \( d = \frac{500}{1.6 \times 10^7} \) and \( C = 0.0000016 \, F \), substitute these into the equation for capacitance \( C = \frac{(8.85 \times 10^{-12} \times 2.6) A}{d} \). 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 \[ u = \frac{1}{2} \varepsilon E^2 \]where \( \varepsilon \) 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 \( \varepsilon_r \) of the material, with the formula:
\[ C = \frac{\varepsilon_0 \varepsilon_r A}{d} \]
where \( \varepsilon_0 \) 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

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, \(\textbf{what is the charge per unit length}\) \(\lambda\) on the capacitor?

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?

A cylindrical air capacitor of length 15.0 m stores 3.20 \(\times\) 10\(^{-9}\) J of energy when the potential difference between the two conductors is 4.00 V. (a) Calculate the magnitude of the charge on each conductor. (b) Calculate the ratio of the radii of the inner and outer conductors.

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?

A parallel-plate air capacitor is to store charge of magnitude 240.0 pC on each plate when the potential difference between the plates is 42.0 V. (a) If the area of each plate is 6.80 cm\(^2\), what is the separation between the plates? (b) If the separation between the two plates is double the value calculated in part (a), what potential difference is required for the capacitor to store charge of magnitude 240.0 pC on each plate?

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