Question

A parallel plate capacitor has a capacitance of \(120 . \mathrm{pF}\) and a plate area of \(100 . \mathrm{cm}^{2}\). The space between the plates is filled with mica whose dielectric constant is \(5.40 .\) The plates of the capacitor are kept at \(50.0 \mathrm{~V}\) a) What is the strength of the electric field in the mica? b) What is the amount of free charge on the plates? c) What is the amount of charge induced on the mica?

Short answer

Answer: The strength of the electric field within the mica is approximately \(1.26 \times 10^5 V/m\). The free charge on the plates is about \(6.0 \times 10^{-9} C\), and the induced charge on the mica is approximately \(4.89 \times 10^{-9} C\).

Step-by-step solution

Calculate the distance between the plates

First, we need to calculate the distance between the plates, \(d\). Using the capacitance formula, we can rearrange it to solve for \(d\): \(d = \frac{k \epsilon_0 A}{C}\) Plug in the given values: \(d = \frac{5.4 \times 8.85 \times 10^{-12} F/m \times 100 \times 10^{-4} m^2}{120 \times 10^{-12} F}\) \(d \approx 3.974 \times 10^{-4} m\)

Calculate the electric field in the mica

Next, we need to find the strength of the electric field in the mica using the formula \(E = \frac{V}{d}\): \(E = \frac{50.0 V}{3.974 \times 10^{-4} m}\) \(E \approx 1.26 \times 10^5 V/m\)

Calculate the free charge on the plates

Now, let's calculate the amount of free charge on the plates using the formula \(Q_{free} = C \times V\): \(Q_{free} = 120 \times 10^{-12} F \times 50.0 V\) \(Q_{free} = 6.0 \times 10^{-9} C\)

Calculate the charge induced on the mica

Finally, we need to find the induced charge on the mica. First, we need to find the charge on the plates without the mica, \(Q_0\). For this, we can use the same capacitance formula but with a dielectric constant of 1: \(C_0 = \frac{\epsilon_0 A}{d}\) Now, find the initial charge on the plates: \(Q_{0} = C_0 \times V = \frac{\epsilon_0 A}{d} \times V\) Replace variables and values: \(Q_{0} = \frac{8.85 \times 10^{-12} F/m \times 100 \times 10^{-4} m^2}{3.974 \times 10^{-4} m} \times 50.0 V\) \(Q_{0} \approx 1.11 \times 10^{-9} C\) Now, calculate the induced charge on the mica: \(Q_{induced} = Q_{free} - Q_{0} = 6.0 \times 10^{-9} C - 1.11 \times 10^{-9} C\) \(Q_{induced} = 4.89 \times 10^{-9} C\)

Key concepts

Capacitance Calculation

Capacitance is a measure of a capacitor's ability to store charge per unit voltage applied across its plates. The formula to calculate the capacitance, (C), of a parallel plate capacitor is:
\( C = \frac{k \epsilon_0 A}{d} \)
where \( k \) is the dielectric constant, \( \epsilon_0 \) is the vacuum permittivity, \( A \) is the plate area, and \( d \) is the separation between the plates. In the context of the exercise, the dielectric constant of mica and the plate area are given. By rearranging the formula to solve for the distance between the plates, we can infer how the dielectric material and plate area impact the overall capacitance. It's critical to appreciate that the capacitance increases with larger plate area, smaller distance, or higher dielectric constant.

Electric Field Strength

The strength of the electric field within a capacitor essentially defines the force per unit charge experienced by a small positive test charge placed between the plates. It is given by the formula:
\( E = \frac{V}{d} \)
where \( E \) stands for electric field strength, \( V \) is the voltage or potential difference applied across the plates, and \( d \) signifies the distance between them. Our exercise demonstrates that by knowing the voltage applied and the distance between the plates, we can determine the uniform field strength inside the capacitor. Higher voltage or closer plates correspond to a stronger electric field. This is pivotal when considering the breakdown voltage of dielectric materials—each has a limit to the electric field it can withstand without becoming conductive.

Dielectric Constant

The dielectric constant, denoted by \( k \), is an essential property of a material that determines its ability to reduce the electric field strength within a capacitor, compared to the vacuum. This directly influences the capacitance since \( k \) appears in the formula for capacitance of a parallel plate capacitor with a dielectric:
\( C = \frac{k \epsilon_0 A}{d} \)
In our example, mica, a dielectric, significantly enhances the capacitance of the capacitor. The dielectric constant of mica is 5.40, which means the presence of mica between the plates allows for more charge to be stored for a given voltage. Understanding the role of the dielectric constant is crucial as it underscores the methodology for selecting appropriate materials in capacitor design to achieve desired properties.

Free and Induced Charge

In a capacitor, the free charge refers to the amount of charge that has been physically deposited on the plates by an external circuit. It is given by the formula:
\( Q_{free} = C \times V \)
where \( Q_{free} \) is the free charge, \( C \) is the capacitance, and \( V \) is the voltage. The induced charge, on the other hand, is the charge that appears on the surface of the dielectric material in response to the electric field produced by the free charge. This phenomenon arises from the polarization of the dielectric. The induced charge affects the overall electric field and therefore the capacitance of the system. By calculating both free and induced charges, as seen in the exercise, students learn the interplay between them and how a dielectric modifies a capacitor's behavior.