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Chapter 19: Problem 47

A parallel plate capacitor has a capacitance of \(7.0 \mu \mathrm{F}\) whenfilled with a dielectric. The area of each plate is \(1.5 \mathrm{m}^{2}\) and the separation between the plates is \(1.0 \times 10^{-5} \mathrm{m} .\) What is the dielectric constant of the dielectric?

Short Answer

Expert verified
The dielectric constant is approximately 5.3.

Step by step solution

01

Identify the known values

We are given the following values: the capacitance \( C = 7.0 \mu \mathrm{F} = 7.0 \times 10^{-6} \mathrm{F} \), the area of each plate \( A = 1.5 \mathrm{m}^2 \), and the separation between the plates \( d = 1.0 \times 10^{-5} \mathrm{m} \).
02

Recall the parallel plate capacitor formula

The capacitance of a parallel plate capacitor is given by \( C = \frac{\kappa \varepsilon_0 A}{d} \), where \( \kappa \) is the dielectric constant, and \( \varepsilon_0 \) is the vacuum permittivity \( (\varepsilon_0 \approx 8.85 \times 10^{-12} \mathrm{F/m}) \).
03

Rearrange the formula to solve for the dielectric constant

The dielectric constant \( \kappa \) can be isolated by rearranging the formula to \( \kappa = \frac{C \cdot d}{\varepsilon_0 \cdot A} \).
04

Substitute the known values into the formula

Plug in the values into the formula: \[ \kappa = \frac{7.0 \times 10^{-6} \mathrm{F} \times 1.0 \times 10^{-5} \mathrm{m}}{8.85 \times 10^{-12} \mathrm{F/m} \times 1.5 \mathrm{m}^2} \].
05

Calculate the result

Simplify and calculate the expression: \[ \kappa = \frac{7.0 \times 10^{-11}}{1.3275 \times 10^{-11}} \approx 5.3 \]. Thus, the dielectric constant of the dielectric is approximately 5.3.

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Key Concepts

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

Parallel Plate Capacitor
A parallel plate capacitor consists of two large, flat plates that are parallel to each other. These plates are usually made of a conductive material and are separated by a small distance filled with a dielectric material. The key purpose of a parallel plate capacitor is to store electrical energy in the electric field that forms between the plates.

Here's how it works:
  • When a voltage is applied across the plates, an electric field is created.
  • This field results from the accumulation of positive charge on one plate and negative charge on the other.
  • The dielectric material between the plates increases the overall capacitance by reducing the field strength for the same charge.
  • The effectiveness of the capacitor largely depends on the surface area of the plates, their distance apart, and the type of dielectric material used in between.
Parallel plate capacitors are commonly used in electronic circuits, providing stability in power supply systems and aiding in signal processing.
Capacitance
Capacitance is a measure of a capacitor's ability to store electrical charge. It is denoted by the symbol \( C \) and is expressed in farads (F). The capacitance of a capacitor depends on various factors such as:
  • The surface area of the plates; larger areas result in greater capacitance.
  • The distance between the plates; closer plates increase the capacitance.
  • The dielectric material; materials with a higher dielectric constant \( \kappa \) result in higher capacitance.
The relation that defines capacitance for a parallel plate capacitor is given by the formula:
\[ C = \frac{\kappa \varepsilon_0 A}{d} \]
where:
  • \( C \) is the capacitance.
  • \( \kappa \) is the dielectric constant of the material between the plates.
  • \( \varepsilon_0 \) is the vacuum permittivity.
  • \( A \) is the area of one of the plates.
  • \( d \) is the separation distance between the plates.
Understanding capacitance is vital for designing circuits that require specific charge and voltage regulation.
Vacuum Permittivity
Vacuum permittivity, represented by \( \varepsilon_0 \), is a physical constant that describes how electric fields interact in a vacuum. It is sometimes referred to as the "dielectric constant of free space" and is crucial in the laws governing electromagnetism.

The value of vacuum permittivity is approximately \( 8.85 \times 10^{-12} \mathrm{F/m} \).
Here are some important points about vacuum permittivity:
  • It is a measure of the ability of the vacuum to permit electric field lines.
  • Plays a critical role in the calculation of capacitor circuits, especially in determining the capacitance of a parallel plate capacitor without a dielectric material.
  • In formulas, it acts as a base reference to understand how much better or worse different dielectrics are in influencing capacitance compared to a vacuum.
Vacuum permittivity provides a foundational understanding needed for deeper study into electric fields and how different materials can influence these fields.

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Most popular questions from this chapter

A positive charge \(q_{1}\) is located \(3.00 \mathrm{m}\) to the left of a negative charge \(q_{2} .\) The charges have different magnitudes. On the line through the charges, the net electric field is zero at a spot \(1.00 \mathrm{m}\) to the right of the negative charge. On this line there are also two spots where the total electric potential is zero. Locate these two spots relative to the negative charge.

A spark plug in an automobile engine consists of two metal conductors that are separated by a distance of \(0.75 \mathrm{mm} .\) When an electric spark jumps between them, the magnitude of the electric field is \(4.7 \times 10^{7} \mathrm{V} / \mathrm{m} .\) What is the magnitude of the potential difference \(\Delta V\) between the conductors?

Two identical capacitors store different amounts of energy: capacitor A stores \(3.1 \times 10^{-3} \mathrm{J},\) and capacitor \(\mathrm{B}\) stores \(3.4 \times 10^{-4} \mathrm{J} .\) The voltage across the plates of capacitor \(\mathrm{B}\) is \(12 \mathrm{V}\). Find the voltage across the plates of capacitor A.

Three point charges, \(-5.8 \times 10^{-9} \mathrm{C},-9.0 \times 10^{-9} \mathrm{C},\) and \(+7.3 \times 10^{-9} \mathrm{C},\) are fixed at different positions on a circle. The total electric potential at the center of the circle is \(-2100 \mathrm{V}\). What is the radius of the circle?

A capacitor is constructed of two concentric conducting cylindrical shells. The radius of the inner cylindrical shell is \(2.35 \times 10^{-3} \mathrm{m}\) and that of the outer shell is \(2.50 \times 10^{-3} \mathrm{m} .\) When the cylinders carry equal and opposite charges of magnitude \(1.7 \times 10^{-10} \mathrm{C},\) the electric field between the plates has an average magnitude of \(4.2 \times 10^{4} \mathrm{V} / \mathrm{m}\) and is directed radially outward from the inner shell to the outer shell. Determine (a) the magnitude of the potential difference between the cylindrical shells and (b) the capacitance of this capacitor.
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