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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.

Short Answer

Expert verified
(a) Charge on plates is \( 5.01 \times 10^{-11} \ \text{C} \). (b) Electric field inside Teflon is \( 1.19 \times 10^{6} \ \text{N/C} \). (c) Electric field without Teflon is \( 2.52 \times 10^{6} \ \text{N/C} \).

Step by step solution

01

Calculate Capacitance with Dielectric

First, let's calculate the capacitance of the parallel-plate capacitor using the formula for a capacitor with a dielectric: \( C = \frac{\varepsilon_r \varepsilon_0 A}{d} \). Here, \( \varepsilon_r \) is the relative permittivity of Teflon (about 2.1), \( \varepsilon_0 \) is the vacuum permittivity (\( 8.85 \times 10^{-12} \ \text{F/m}\)), \( A = 0.0225 \ \text{m}^2 \), and \( d = 1.00 \times 10^{-3} \ \text{m} \). Calculate \( C \):\[ C = \frac{2.1 \times 8.85 \times 10^{-12} \times 0.0225}{1.00 \times 10^{-3}} \approx 4.175 \times 10^{-12} \ \text{F} \]
02

Calculate Charge on Plates

Now that we have the capacitance, we can calculate the charge \( Q \) on the plates using \( Q = CV \), where \( V = 12.0 \ \text{V} \).\[ Q = 4.175 \times 10^{-12} \times 12.0 \approx 5.01 \times 10^{-11} \ \text{C} \]
03

Apply Gauss's Law to Find Electric Field with Dielectric

Gauss's law in terms of electric field \( E \) is \( E = \frac{\sigma}{\varepsilon_0 \varepsilon_r} \), where \( \sigma = \frac{Q}{A} \) and \( \varepsilon_r = 2.1 \):\[ \sigma = \frac{5.01 \times 10^{-11}}{0.0225} \approx 2.23 \times 10^{-9} \ \text{C/m}^2 \]\[ E = \frac{2.23 \times 10^{-9}}{8.85 \times 10^{-12} \times 2.1} \approx 1.19 \times 10^{6} \ \text{N/C} \]
04

Calculate Electric Field Without Dielectric

If the dielectric is removed, the electric field is determined just by \( E = \frac{\sigma}{\varepsilon_0} \).\[ E = \frac{2.23 \times 10^{-9}}{8.85 \times 10^{-12}} \approx 2.52 \times 10^{6} \ \text{N/C} \]

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Key Concepts

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

Electric Field
The electric field is a fundamental concept in physics that describes the force per unit charge at a given point in space. It's essentially a vector field that exerts force on electric charges. In simpler terms, any charge placed in an electric field will experience a force that acts on it. The strength and direction of this force depend on both the magnitude of the field and the amount of charge present.

In the context of a capacitor, the electric field between the plates can be calculated using Gauss's law. The electric field parallels the line perpendicular to the plates because the field lines emanate directly between them. For a parallel-plate capacitor, the electric field (E) is uniformly distributed and is given by the equation:\[E = \frac{\sigma}{\varepsilon_0}\] where \(\sigma\) is the charge density, and \(\varepsilon_0\) is the permittivity of free space.

Understanding the electric field helps visualize how capacitors store energy and influence charges placed in the field. It is vital to remember that the presence of dielectric materials, like Teflon in our problem, influences the electric field, altering its magnitude by a factor equal to the relative permittivity of the dielectric.
Gauss's Law
Gauss's law is a powerful tool for analyzing electric fields, especially in cases with high symmetry, like spherical, cylindrical, or planar symmetry. This law states that the total electric flux through a closed surface is equal to the total charge enclosed divided by the permittivity of free space. Mathematically, Gauss's law is expressed as:\[\oint \mathbf{E} \cdot d\mathbf{A} = \frac{Q_{\text{enc}}}{\varepsilon_0}\]This equation illustrates that the electric field lines entering and exiting a given surface depend directly on the charge inside the surface.

In a parallel-plate capacitor scenario, Gauss's law simplifies to describe the relationship between the charge density, electric field, and any dielectric present. When a dielectric is present, the equation modifies to:\[E = \frac{\sigma}{\varepsilon_0 \varepsilon_r}\]where \(\varepsilon_r\) represents the dielectric's relative permittivity.

This aspect of Gauss's law helps calculate the electric field strength inside the capacitor, considering both dielectric material and charge distribution. It gives insights into how the field and flux relate to the physical properties of the system, making it a vital consideration when analyzing capacitors with and without dielectric materials.
Capacitance with Dielectric
Capacitance is the ability of a capacitor to store energy in the form of charge for a given voltage. Adding a dielectric material between capacitor plates is a common way to enhance capacitance. Dielectrics increase the overall capacitance by reducing the electric field for the same amount of charge on the plates. This occurs because the dielectric material polarizes in the field, effectively lowering the space's electric field strength.

The capacitance of a parallel-plate capacitor with a dielectric is given by:\[C = \frac{\varepsilon_r \varepsilon_0 A}{d}\]where \(\varepsilon_r\) is the dielectric's relative permittivity, \(A\) is the plate area, and \(d\) is the distance between the plates.

The dielectric constant \(\varepsilon_r\) is crucial because it indicates how much the dielectric increases the capacitance compared to a vacuum gap. For example, in our exercise, we use Teflon with a relative permittivity of about 2.1. This means the Teflon allows the capacitor to store more charge at a given voltage than if there were no dielectric. Knowing how dielectric affects capacitance helps in designing more efficient capacitor systems for various electrical and electronic applications.

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Most popular questions from this chapter

A parallel-plate capacitor has capacitance \(C_0\) = 8.00 pF when there is air between the plates. The separation between the plates is 1.50 mm. (a) What is the maximum magnitude of charge Q that can be placed on each plate if the electric field in the region between the plates is not to exceed 3.00 \(\times\) 10\(^4\) V/m? (b) A dielectric with \(K = 2.70\) is inserted between the plates of the capacitor, completely filling the volume between the plates. Now what is the maximum magnitude of charge on each plate if the electric field between the plates is not to exceed 3.00 \(\times\) 10\(^4\) V/m?

A parallel-plate vacuum capacitor has 8.38 J of energy stored in it. The separation between the plates is 2.30 mm. If the separation is decreased to 1.15 mm, what is the energy stored (a) if the capacitor is disconnected from the potential source so the charge on the plates remains constant, and (b) if the capacitor remains connected to the potential source so the potential difference between the plates remains constant?

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?

An air capacitor is made by using two flat plates, each with area A, separated by a distance \(d\). Then a metal slab having thickness \(a\) (less than \(d\)) and the same shape and size as the plates is inserted between them, parallel to the plates and not touching either plate (\(\textbf{Fig. P24.62}\)). (a) What is the capacitance of this arrangement? (b) Express the capacitance as a multiple of the capacitance C\(_0\) when the metal slab is not present. (c) Discuss what happens to the capacitance in the limits \(a\) \(\rightarrow\) 0 and \(a\) \(\rightarrow\) d.

A parallel-plate air capacitor is to store charge of magnitude 240.0 pC on each plate when the potential difference between the plates is 42.0 V. (a) If the area of each plate is 6.80 cm\(^2\), what is the separation between the plates? (b) If the separation between the two plates is double the value calculated in part (a), what potential difference is required for the capacitor to store charge of magnitude 240.0 pC on each plate?

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