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Chapter 25: Problem 27

A parallel-plate capacitor has a capacitance of \(100 \mathrm{pF}\), a plate area of \(80 \mathrm{~cm}^{2}\), and a mica dielectric \((\kappa=5.4)\) completely filling the space between the plates. At \(85 \mathrm{~V}\) potential difference, calculate (a) the electric field magnitude \(E\) in the mica, (b) the magnitude of the free charge on the plates, and (c) the magnitude of the induced surface charge on the mica.

Short Answer

Expert verified
Electric field \(E\) is approximately 222,193 V/m; free charge \(Q\) is 8.5 nC; induced charge \(Q_i\) is 7 nC.

Step by step solution

01

Identify Known Values

We are given that the capacitance \(C\) is 100 pF, the plate area \(A\) is 80 cm\(^2\), the dielectric constant \(\kappa\) is 5.4, and the potential difference \(V\) is 85 V.
02

Calculate the Electric Field (E) in the Mica

The electric field \(E\) in a capacitor is given by the formula \(E = \frac{V}{d}\), where \(d\) is the separation between the plates. However, since we do not have \(d\), we use the relationship \(C = \frac{\kappa \varepsilon_0 A}{d}\) to find it. Rearranging gives \(d = \frac{\kappa \varepsilon_0 A}{C}\).The permittivity of free space \(\varepsilon_0\) is approximately \(8.85 \times 10^{-12} \mathrm{F/m}\). First, convert 80 cm\(^2\) to m\(^2\): \[A = 80 \times 10^{-4} \text{ m}^2\]Now solve for \(d\):\[d = \frac{5.4 \times 8.85 \times 10^{-12} \times 80 \times 10^{-4}}{100 \times 10^{-12}} \approx 3.83 \times 10^{-4} \text{ m}\]Now calculate \(E\):\[E = \frac{85}{3.83 \times 10^{-4}} \approx 222,193 \text{ V/m}\]
03

Calculate the Free Charge (Q) on the Plates

The free charge \(Q\) on the plates can be calculated using the formula \(Q = CV\):\[Q = 100 \times 10^{-12} \times 85 = 8.5 \times 10^{-9} \text{ C}\]
04

Calculate the Induced Surface Charge on the Mica

The induced surface charge on the mica \(Q_i\) is given by the formula \(Q_i = Q \left(1 - \frac{1}{\kappa}\right)\):\[Q_i = 8.5 \times 10^{-9} \times \left(1 - \frac{1}{5.4}\right) \approx 7 \times 10^{-9} \text{ C}\]

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Key Concepts

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

Capacitance
In a parallel-plate capacitor, the concept of capacitance is fundamental. Capacitance is the ability of a system to store an electric charge. The unit of capacitance is the farad (F), which denotes how much charge can be held per volt of electric potential difference across the capacitor's plates. In this context, the capacitor's capacitance is determined by:
  • The area of the plates, denoted as \(A\).
  • The separation between the plates, \(d\), which influences the electric field.
  • The dielectric material between the plates which determines the dielectric constant \(\kappa\).
The relationship between these factors is expressed by the formula \(C = \frac{\kappa \varepsilon_0 A}{d}\), where \(\varepsilon_0\) is the permittivity of free space. By utilizing the given values, we understand how the capacitor's structure and materials influence its ability to hold and manage charge.
Dielectric Constant
The dielectric constant, \(\kappa\), is a measure of a material's ability to increase the capacitance of a capacitor compared to if it were filled with a vacuum. This constant is crucial when determining how effectively a material can store electrical energy.
  • A higher dielectric constant means more charge can be stored at the same voltage.
  • Mica, the dielectric in our exercise, has a dielectric constant of 5.4, indicating significant capacitive enhancement.
In our problem, the dielectric constant not only influences the capacitance but also affects the electric field within the capacitor. This makes the dielectric constant an essential factor in determining the capacitor's efficiency and overall performance.
Electric Field Calculation
The electric field \(E\) in a capacitor is the force per unit charge between the plates, often visualized as the lines of force set up by the separated charges. For a parallel-plate capacitor, the electric field is calculated using:
  • \(E = \frac{V}{d}\), the voltage \(V\) across the plates divided by the distance \(d\) between them.
Since \(d\) is derived from the relationship \(C = \frac{\kappa \varepsilon_0 A}{d}\), we rearrange to find this distance, enabling the calculation of \(E\). Our calculation yields an electric field magnitude of approximately 222,193 V/m within the mica, confirming how the dielectric affects the electric potential energy distribution across the capacitor.
Surface Charge
Surface charge relates to the charge per area on the plates and the induced charge within the dielectric. When a dielectric is introduced, two essential types of charges are considered:
  • Free charge \(Q\) - This is the charge directly associated with the capacitor's plates, given by \(Q = CV\).
  • Induced surface charge \(Q_i\) - The dielectric polarizes in response to the electric field, creating internal charges that reduce the external field effect.
The formula \(Q_i = Q \left(1 - \frac{1}{\kappa}\right)\) helps us calculate these polarized, or induced, charge components, which are critical for understanding the overall interaction within the capacitor. This concept of surface charge is significant in defining the effectiveness and limits of dielectric materials in capacitive applications.

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Most popular questions from this chapter

A \(100 \mathrm{pF}\) capacitor is charged to a potential difference of \(80.0 \mathrm{~V}\), and the charging battery is disconnected. The capacitor is then connected in parallel with a second (initially uncharged) capacitor. If the potential difference across the first capacitor drops to \(35.0 \mathrm{~V}\), what is the capacitance of this second capacitor?

Assume that a stationary electron is a point of charge. What is the energy density \(u\) of its electric field at radial distances (a) \(r=1.00 \mathrm{~mm}\), (b) \(r=1.00 \mu \mathrm{m}\), (c) \(r=1.00 \mathrm{~nm}\), (d) \(r=1.00 \mathrm{pm}\), and (e) \(r=1.00 \mathrm{fm}\) ? (f) What is \(u\) in the limit as \(r \rightarrow 0\) ?

You have two flat metal plates, each of area \(1.00 \mathrm{~m}^{2}\), with which to construct a parallel-plate capacitor. (a) If the capacitance of the device is to be \(2.00 \mathrm{~F}\), what must be the separation between the plates? (b) Could this capacitor actually be constructed?

Figure 25-30 shows a parallel-plate capacitor with a plate area \(A=7.89 \mathrm{~cm}^{2}\) and plate separation \(d=4.62 \mathrm{~mm}\). The top half of the gap is filled with material of dielectric constant \(\kappa_{1}=11.0\); the bottom half is filled with material of dielectric constant \(\kappa_{2}=4.0\). What is the capacitance?

A coaxial cable used in a transmission line has an inner radius of \(0.10 \mathrm{~mm}\) and an outer radius of \(0.40 \mathrm{~mm}\). Calculate the capacitance per meter for the cable. Assume that the space between the conductors is filled with epoxy resin with dielectric constant \(3.6\).
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