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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

Identify the Formula

The capacitance of a parallel plate capacitor with a dielectric can be calculated using the formula: \[ C = \frac{\varepsilon_r \cdot \varepsilon_0 \cdot A}{d} \]where \(C\) is the capacitance, \(\varepsilon_r\) is the dielectric constant, \(\varepsilon_0\) is the permittivity of free space \(8.85 \times 10^{-12} \, \text{F/m}\), \(A\) is the area of the plates, and \(d\) is the separation between the plates.
02

Rearrange Formula to solve for Dielectric Constant

To find the dielectric constant \(\varepsilon_r\), rearrange the formula:\[ \varepsilon_r = \frac{C \cdot d}{\varepsilon_0 \cdot A} \]
03

Plug in the Known Values

Insert the given values into the formula:- \(C = 7.0 \, \mu\text{F} = 7.0 \times 10^{-6} \, \text{F}\)- \(d = 1.0 \times 10^{-5} \, \text{m}\)- \(\varepsilon_0 = 8.85 \times 10^{-12} \, \text{F/m}\)- \(A = 1.5 \, \text{m}^2\)Then\[ \varepsilon_r = \frac{7.0 \times 10^{-6} \cdot 1.0 \times 10^{-5}}{8.85 \times 10^{-12} \cdot 1.5} \]
04

Perform the Calculations

Calculate the dielectric constant:\[ \varepsilon_r = \frac{7.0 \times 10^{-11}}{13.275 \times 10^{-12}} \approx 5.27 \]
05

Conclusion

The dielectric constant of the dielectric is approximately 5.27.

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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 is an essential component used frequently in electronic circuits. It consists of two conductive plates separated by a distance. The plates are parallel to each other, hence the name, "parallel plate capacitor." These plates store electric charge, which is crucial for various applications such as filtering and storing energy in power supplies.

The capacitance (\(C\)) of a parallel plate capacitor indicates how much charge it can store per unit voltage. It's measured in farads (F) and is influenced by three main factors:
  • The area (\(A\)) of the plates, where larger plates can store more charges, increasing capacitance.
  • The separation distance (\(d\)) between the plates, where the closer the plates, the higher the capacitance.
  • The dielectric material between the plates, which can alter the capacitor’s overall capacitance.


In practice, parallel plate capacitors are straightforward and efficient for basic setups, especially when used with direct and alternating current electronics.
capacitance formula
To understand how capacitance works, it's important to know the formula governing a parallel plate capacitor:\[C = \frac{\varepsilon_0 \cdot A}{d}\] This formula adjusts when a dielectric material is present between the plates:\[C = \frac{\varepsilon_r \cdot \varepsilon_0 \cdot A}{d}\] Here,
  • \(C\) is the capacitance in farads (F).
  • \(\varepsilon_0\) represents the permittivity of free space (\(8.85 \times 10^{-12} \, \text{F/m}\)), a fundamental constant.
  • \(A\) is the area of a single plate in square meters (\(\text{m}^2\)).
  • \(d\) is the distance between the two plates in meters (m).
  • \(\varepsilon_r\) is the dielectric constant, a multiplier added when a dielectric material is used, indicating how much more charge the capacitor can store compared to having nothing between its plates.


This formula reveals that the capacitance increases with a larger plate area, high dielectric constant, and reduced distance between the plates. Knowing this formula helps you tweak capacitors for desired outcomes in electronic circuits.
dielectrics in capacitors
Dielectrics are materials inserted between the plates of a capacitor to enhance its capacity to store electrical charge. Normally, air is the medium between plates, but inserting a dielectric increases the stored charge without changing the physical attributes of the capacitor. This happens because dielectrics reduce the electric field inside the capacitor, allowing more charge to be stored for the same voltage.

The dielectric constant (\(\varepsilon_r\)) quantifies the effectiveness of a dielectric material. It's a dimensionless number signifying how much more electrical charge a capacitor can store compared to having a vacuum between its plates, where the dielectric constant would be 1. When calculating the capacitance of a capacitor with a dielectric, the constant serves as a critical element in the formula, amplifying the results.

The benefits of using dielectrics in capacitors contain:
  • Increased capacitance, due to the greater ability to store charge.
  • Improved insulation between the plates, minimizing the risk of voltage breakdown.
  • Versatility in electronic design, as engineers can choose from various dielectric materials to tailor the capacitors’ properties.

Overall, the integration of dielectrics allows for more efficient and effective capacitor designs, making them indispensable in modern electronics.

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Most popular questions from this chapter

An electric field has a constant value of \(4.0 \times 10^{3} \mathrm{V} / \mathrm{m}\) and is directed downward. The field is the same everywhere. The potential at a point \(P\) within this region is 155 V. Find the potential at the following points: (a) \(6.0 \times 10^{-3} \mathrm{m}\) directly above \(P,(\) (b) \(3.0 \times 10^{-3} \mathrm{m}\) directly below \(P\) (c) \(8.0 \times 10^{-3} \mathrm{m}\) directly to the right of \(P\)

A particle with a charge of \(-1.5 \mu \mathrm{C}\) and a mass of \(2.5 \times 10^{-6} \mathrm{kg}\) is released from rest at point \(A\) and accelerates toward point \(B\), arriving there with a speed of \(42 \mathrm{m} / \mathrm{s} .\) The only force acting on the particle is the electric force. (a) Which point is at the higher potential? Give your reasoning. (b) What is the potential difference \(V_{\mathrm{B}}-V_{\mathrm{A}}\) between \(\mathrm{A}\) and \(\mathrm{B} ?\)

A charge of \(-3.00 \mu \mathrm{C}\) is fixed in place. From a horizontal distance of \(0.0450 \mathrm{m},\) a particle of \(\operatorname{mass} 7.20 \times 10^{-3} \mathrm{kg}\) and charge \(-8.00 \mu \mathrm{C}\) is fired with an initial speed of \(65.0 \mathrm{m} / \mathrm{s}\) directly toward the fixed charge. How far does the particle travel before its speed is zero?

An electron and a proton are initially very far apart (effectively an infinite distance apart). They are then brought together to form a hydrogen atom, in which the electron orbits the proton at an average distance of \(5.29 \times\) \(10^{-11} \mathrm{m} .\) What is \(\mathrm{EPE}_{\text {final }}-\mathrm{EPE}_{\text {initial }},\) which is the change in the electric potential energy?

One particle has a mass of \(3.00 \times 10^{-3} \mathrm{kg}\) and a charge of \(+8.00 \mu \mathrm{C}\). A second particle has a mass of \(6.00 \times 10^{-3} \mathrm{kg}\) and the same charge. The two particles are initially held in place and then released. The particles fly apart, and when the separation between them is \(0.100 \mathrm{m},\) the speed of the \(3.00 \times 10^{-3} \mathrm{kg}\) particle is \(125 \mathrm{m} / \mathrm{s} .\) Find the initial separation between the particles.

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