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A parallel plate capacitor has a capacitance of 7.0\(\mu \mathrm{F}\) when filled 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 5.27.

Step by step solution

01

Understand the formula for capacitance

The capacitance of a parallel plate capacitor filled with a dielectric is given by the formula: \( C = \frac{\kappa \varepsilon_0 A}{d} \), where \( C \) is the capacitance, \( \kappa \) is the dielectric constant, \( \varepsilon_0 \) is the vacuum permittivity \( (8.85 \times 10^{-12} \ \mathrm{F/m}) \), \( A \) is the area of one plate, and \( d \) is the separation between the plates.
02

Rearrange the formula to solve for the dielectric constant

We need to find the dielectric constant \( \kappa \). Rearrange the formula to get \( \kappa = \frac{C \cdot d}{\varepsilon_0 \cdot A} \). This will allow us to solve for \( \kappa \) once we substitute in the given values.
03

Substitute the known values into the formula

Insert the known values into the formula: \( C = 7.0 \times 10^{-6} \ \mathrm{F} \), \( A = 1.5 \ \mathrm{m^2} \), \( d = 1.0 \times 10^{-5} \ \mathrm{m} \), and \( \varepsilon_0 = 8.85 \times 10^{-12} \ \mathrm{F/m} \). Thus the equation becomes \( \kappa = \frac{7.0 \times 10^{-6} \times 1.0 \times 10^{-5}}{8.85 \times 10^{-12} \times 1.5} \).
04

Perform the calculation

Compute the value: \( \kappa = \frac{7.0 \times 10^{-11}}{1.3275 \times 10^{-11}} \approx 5.27 \). Thus, the dielectric constant of the material is approximately 5.27.

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Key Concepts

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

Dielectric constant calculation
When working with a parallel plate capacitor, understanding how to calculate the dielectric constant, represented by \( \kappa \), is crucial. The dielectric constant is a measure of a material's ability to increase the capacitance of the capacitor it fills.
It tells us how much more effective a capacitor is with the dielectric material compared to a vacuum. To calculate the dielectric constant, you use the formula:
  • \( \kappa = \frac{C \cdot d}{\varepsilon_0 \cdot A} \)
Here:
  • \( C \) is the capacitance of the capacitor when filled with the dielectric (in Farads).
  • \( d \) refers to the distance between the parallel plates (in meters).
  • \( \varepsilon_0 \) is the vacuum permittivity, a constant value.
  • \( A \) represents the area of one of the plates (in square meters).
By substituting the known values into this equation, you can determine the dielectric constant, which, in this exercise, was approximately 5.27. This calculation shows how the dielectric affects the capacitor's performance.
Parallel plate capacitor formula
The parallel plate capacitor formula is fundamental in calculating the attributes of a capacitor consisting of two parallel conductive plates. This arrangement is widely used due to its simplicity and effectiveness. The standard formula to compute the capacitance \( C \) is:
  • \( C = \frac{\kappa \varepsilon_0 A}{d} \)
Key elements of this formula include:
  • The dielectric constant \( \kappa \) that represents how the material between the plates changes the electric field and charges storage capacity.
  • The vacuum permittivity \( \varepsilon_0 \), a constant denoting how electric field propagates in a vacuum.
  • The area \( A \) of the plates, which indicates the available space for charge.
  • The plate separation \( d \), affecting how much electric field can build up between the plates.
By properly arranging these factors, you can efficiently calculate and optimize a capacitor's characteristics for various electronic applications.
Vacuum permittivity
Vacuum permittivity, often represented by \( \varepsilon_0 \), is a constant that characterizes the electric force interaction in a vacuum. It is a critical component in the formulas used to calculate capacitance which shows how well a vacuum can permit electric field lines.
In the International System of Units (SI), it's valued at \( 8.85 \times 10^{-12} \ \, \mathrm{F/m} \) (farads per meter). This constant is essential because it provides a baseline for comparing the electric field's effectiveness in different materials. When a dielectric material fills a capacitor, the \( \varepsilon_0 \) in the formula gets adjusted by the dielectric constant \( \kappa \), indicating how much better the material is at permitting electric field lines compared to a vacuum. Every parallel plate capacitor's performance relies heavily on this constant, making it indispensable for understanding and designing effective capacitance components in electronic circuits.

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Most popular questions from this chapter

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