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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 parallel-plate capacitor has plates with area 0.0225 m\(^2\) separated by 1.00 mm of Teflon. (a) Calculate the charge on the plates when they are charged to a potential difference of 12.0 V. (b) Use Gauss's law (Eq. 24.23) to calculate the electric field inside the Teflon. (c) Use Gauss's law to calculate the electric field if the voltage source is disconnected and the Teflon is removed.

Two parallel plates have equal and opposite charges. When the space between the plates is evacuated, the electric field is \(E = 3.20\times10^5\space V/m\). When the space is filled with dielectric, the electric field is \(E = 2.50\times10^5\space V/m\). (a) What is the charge density on each surface of the dielectric? (b) What is the dielectric constant?

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?

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?

In one type of computer keyboard, each key holds a small metal plate that serves as one plate of a parallel-plate, airfilled capacitor. When the key is depressed, the plate separation decreases and the capacitance increases. Electronic circuitry detects the change in capacitance and thus detects that the key has been pressed. In one particular keyboard, the area of each metal plate is 42.0 mm\(^2\), and the separation between the plates is 0.700 mm before the key is depressed. (a) Calculate the capacitance before the key is depressed. (b) If the circuitry can detect a change in capacitance of 0.250 pF, how far must the key be depressed before the circuitry detects its depression?

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