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The paper dielectric in a paper-and-foil capacitor is \(0.0800 \mathrm{~mm}\) thick. Its dielectric constant is \(2.50,\) and its dielectric strength is \(50.0 \mathrm{MV} / \mathrm{m} .\) Assume that the geometry is that of a parallel- plate capacitor, with the metal foil serving as the plates. (a) What area of each plate is required for a \(0.200 \mu \mathrm{F}\) capacitor? (b) If the electric field in the paper is not to exceed one-half the dielectric strength, what is the maximum potential difference that can be applied across the capacitor?

Short Answer

Expert verified
(a) 7.23 × 10^{-3} m²; (b) 2000 V.

Step by step solution

01

Calculate the Capacitance Formula

For a parallel-plate capacitor, the capacitance \( C \) is given by the formula \[ C = \frac{\varepsilon_0 \varepsilon_r A}{d} \] where \( \varepsilon_0 \) is the permittivity of free space \( 8.85 \times 10^{-12} \text{ F/m} \), \( \varepsilon_r \) is the dielectric constant (given as 2.50), \( A \) is the area of the plates, and \( d \) is the separation between the plates (given as 0.0800 mm, or 0.0800 \times 10^{-3} \text{ m}).
02

Solve for the Area of Each Plate (a)

Rearrange the capacitance formula to solve for \( A \): \[ A = \frac{C \, d}{\varepsilon_0 \varepsilon_r} \]Substitute \( C = 0.200 \times 10^{-6} \text{ F} \), \( d = 0.0800 \times 10^{-3} \text{ m} \), and compute the area:\[ A = \frac{0.200 \times 10^{-6} \times 0.0800 \times 10^{-3}}{8.85 \times 10^{-12} \times 2.50} \approx 7.23 \times 10^{-3} \text{ m}^2 \]
03

Determine Maximum Electric Field (b)

The dielectric strength is \( 50.0 \text{ MV/m} \), and we are to keep the electric field at half this value:\[ E_{\text{max}} = \frac{50.0 \times 10^6}{2} = 25.0 \times 10^6 \text{ V/m} \]
04

Calculate Maximum Potential Difference (b)

Use \( V = E \times d \) to find the maximum potential difference, substituting \( E = 25.0 \times 10^6 \text{ V/m} \) and \( d = 0.0800 \times 10^{-3} \text{ m} \):\[ V_{\text{max}} = 25.0 \times 10^6 \times 0.0800 \times 10^{-3} = 2000 \text{ V} \]

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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 the relative permittivity, is a measure of the ability of a material to store electrical energy within an electric field. It is denoted by the symbol \( \varepsilon_r \). The dielectric constant is a dimensionless quantity and provides a comparison to vacuum, which has a value of 1. Essentially, it shows how much more effectively a material can hold electric charge compared to a vacuum.
  • A higher dielectric constant means the material can store more charge.
  • The dielectric constant of paper, in this case, is 2.50, meaning it's 2.5 times better at storing energy than vacuum.
This property is crucial when designing capacitors, as it directly influences the capacitance according to the formula \( C = \frac{\varepsilon_0 \varepsilon_r A}{d} \), where \( C \) is the capacitance, \( \varepsilon_0 \) is the permittivity of free space, \( A \) is the area of the plates, and \( d \) is the separation between the plates.
Dielectric Strength
Dielectric strength represents the maximum electric field a material can withstand without breaking down or losing its insulating properties. It is expressed in units of voltage per unit thickness, often volts per meter (V/m). In the given example, the dielectric strength of the paper is 50.0 MV/m. This means it can endure an electric field up to this value before it begins to conduct electricity, risking damage or failure.
  • Dielectric strength is critical for capacitors as it determines the upper limit on potential difference or voltage that can be safely applied.
  • In practice, to ensure safety and longevity, capacitors are often operated below their absolute dielectric strength.
By keeping the electric field below a certain level, such as half in this example, we can apply a voltage up to 2000 V without surpassing the safe operational limit.
Parallel-Plate Capacitor
A parallel-plate capacitor is a basic capacitor structure consisting of two conductive plates separated by a dielectric material. The capacitor's ability to store electrical energy is largely determined by the surface area of the plates, the separation distance between them, and the dielectric material used.
  • Capacitance \( C \) increases with larger plate area \( A \), as more area allows more charge storage.
  • The separation \( d \) inversely affects capacitance; a larger gap decreases capacitance because the electric field strength needed to keep charges separated grows.
In this context, with a dielectric constant material between the plates, the formula \( C = \frac{\varepsilon_0 \varepsilon_r A}{d} \) helps calculate the desired capacitance by adjusting area and material properties accordingly. This formula was used to find the necessary area of the plates to meet the specified capacitance of 0.200 \( \mu \)F.
Permittivity of Free Space
Permittivity of free space, denoted by \( \varepsilon_0 \), is a fundamental physical constant that characterizes the ability of a vacuum to permit electric field lines. It is essential in the formulation of many electromagnetic equations and appears frequently in the equations governing capacitors. The permittivity of free space has a value of approximately \( 8.85 \times 10^{-12} \) F/m (farads per meter).
  • This constant is key in determining the capacitance of a capacitor; you will see how it integrates with the dielectric constant \( \varepsilon_r \) and geometry to influence the final capacitance.
  • In our specific problem, using this constant helps quantify the baseline ability of the area between the conductive plates to store charge in the absence of any dielectric material.
Knowing \( \varepsilon_0 \) allows us to understand and calibrate how additional materials enhance the capacitance when placed between the plates to amplify the storage capabilities of a capacitor.

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Most popular questions from this chapter

Two large metal parallel plates carry opposite charges of equal magnitude. They are separated by \(45.0 \mathrm{~mm},\) and the potential difference between them is \(360 \mathrm{~V}\). (a) What is the magnitude of the electric field (assumed to be uniform) in the region between the plates? (b) What is the magnitude of the force this field exerts on a particle with charge \(+2.40 \mathrm{nC} ?\)

A capacitor consists of two parallel plates, each with an area of \(16.0 \mathrm{~cm}^{2},\) separated by a distance of \(0.200 \mathrm{~cm} .\) The material that fills the volume between the plates has a dielectric constant of \(5.00 .\) The plates of the capacitor are connected to a \(300 \mathrm{~V}\) battery. (a) What is the capacitance of the capacitor? (b) What is the charge on either plate? (c) How much energy is stored in the charged capacitor?

A disk-shaped parallel-plate capacitor has a capacitance \(C\). In terms of \(C,\) what would the capacitance be if the radius of each plate were increased by a factor of \(3 ?\)

(a) An electron is to be accelerated from \(3.00 \times 10^{6} \mathrm{~m} / \mathrm{s}\) to \(8.00 \times 10^{6} \mathrm{~m} / \mathrm{s} .\) Through what potential difference must the electron pass to accomplish this? (b) Through what potential difference must the electron pass if it is to be slowed from \(8.00 \times 10^{6} \mathrm{~m} / \mathrm{s}\) to a halt?

A charge of \(28.0 \mathrm{nC}\) is placed in a uniform electric field that is directed vertically upward and that has a magnitude of \(4.00 \times 10^{4} \mathrm{~N} / \mathrm{C} .\) What work is done by the electric force when the charge moves (a) \(0.450 \mathrm{~m}\) to the right; (b) \(0.670 \mathrm{~m}\) upward; (c) \(2.60 \mathrm{~m}\) at an angle of \(45.0^{\circ}\) downward from the horizontal?

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