Question

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

(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

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 \( C \):\[ C = \frac{2.1 \times 8.85 \times 10^{-12} \times 0.0225}{1.00 \times 10^{-3}} \approx 4.175 \times 10^{-12} \ \text{F} \]

Calculate Charge on Plates

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

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 \( \varepsilon_r = 2.1 \):\[ \sigma = \frac{5.01 \times 10^{-11}}{0.0225} \approx 2.23 \times 10^{-9} \ \text{C/m}^2 \]\[ E = \frac{2.23 \times 10^{-9}}{8.85 \times 10^{-12} \times 2.1} \approx 1.19 \times 10^{6} \ \text{N/C} \]

Calculate Electric Field Without Dielectric

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

Key concepts

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.