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Two parallel plates of area \(100 \mathrm{~cm}^{2}\) are given excess charges of equal amounts \(8.9 \times 10^{-7} \mathrm{C}\) but opposite signs. The electric field within the dielectric material filling the space between the plates is \(1.4 \times 10^{6} \mathrm{~V} / \mathrm{m} .\) (a) Calculate the dielectric constant of the material. (b) Determine the amount of bound charge induced on each dielectric surface.

Short Answer

Expert verified
The dielectric constant is approximately 71788.6 and the bound charge induced on each dielectric surface is approximately 8.9 脳 10鈦烩伖 C.

Step by step solution

01

- Identify given values

First, identify the given values in the problem:- Area of the plates, A = 100 cm虏 = 0.01 m虏 (since 1 cm虏 = 0.0001 m虏)- Charge on plates, Q = 8.9 脳 10鈦烩伔 C- Electric field, E = 1.4 脳 10鈦 V/m
02

- Use the formula for electric field between plates

The electric field between the plates without dielectric is given by the formula: \[ E_0 = \frac{Q}{\varepsilon_0 A} \]where \( E_0 \) is the electric field without dielectric,\( \varepsilon_0 \) is the vacuum permittivity \( (8.85 \times 10^{-12} \mathrm{Fm^{-1}}) \),\( Q \) is the charge, and\( A \) is the area of the plates.Calculate \( E_0 \):\[ E_0 = \frac{8.9 \times 10^{-7}}{8.85 \times 10^{-12} \times 0.01} = 1.005 \times 10^{11} \mathrm{V/m} \]
03

- Apply dielectric constant formula

The dielectric constant \( (\kappa) \) is given by the ratio of the electric field without the dielectric to the electric field with the dielectric:\[ \kappa = \frac{E_0}{E} \]Plug in the values:\[ \kappa = \frac{1.005 \times 10^{11}}{1.4 \times 10^6} = 71788.6 \]
04

- Calculate bound charge density

Bound surface charge density, \( \sigma_b \), on the dielectric is given by:\[ \sigma_b = Q \left(1 - \frac{1}{\kappa} \right)\]Plug in the values:\[ \sigma_b = 8.9 \times 10^{-7} \left(1 - \frac{1}{71788.6} \right) \approx 8.9 \times 10^{-7} \mathrm{C} \]
05

- Calculate induced bound charge

The amount of induced bound charge on each dielectric surface \(Q_b\) can be found:\[ Q_b = \sigma_b \cdot A \]Plug in the values:\[ Q_b = 8.9 \times 10^{-7} \times 0.01 = 8.9 \times 10^{-9} \mathrm{C} \]

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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 (\textbf{E}) is a crucial concept in electrostatics. It quantifies the force per unit charge experienced by a positive test charge in the vicinity of other electric charges. Mathematically, it is expressed as:

\textbf{E} = \frac{\textbf{F}}{\textbf{q}}

where \textbf{F} is the force experienced by the charge \textbf{q}. In the context of parallel plate capacitors, the electric field is uniform between the plates and perpendicular to their surfaces.

When a dielectric material is inserted between the plates, it modifies the electric field. This modified field (\textbf{E}) is weaker than the field without the dielectric (\textbf{E\(_0\)}). The relationship between the fields can be expressed using the dielectric constant (\textbf{魏}), as:

\textbf{魏} = \frac{\textbf{E\(_0\)}}{\textbf{E}}

The dielectric constant, hence, is a measure of how much the dielectric reduces the electric field within it.
Dielectric Materials
Dielectric materials are insulating substances that do not conduct electricity but can be polarized in the presence of an electric field. When placed between the plates of a capacitor, dielectrics reduce the electric field and store energy. These materials increase the capacitor's ability to store charge without increasing the voltage applied.

When exposed to an electric field, dielectric materials exhibit a phenomenon called polarization. This occurs because the electric field causes positive and negative charges within the dielectric to align in opposite directions. As a result, bound charges appear on the surfaces of the dielectric, opposing the applied electric field and thereby reducing the overall field strength.

Using the given problem, we can calculate the dielectric constant (\textbf{魏}) of the material as follows:

\textbf{魏} = \frac{E\(_0\)}{E} = \frac{1.005 \times 10^{11}}{1.4 \times 10^6} \textbf{鈮 71788.6}

This illustrates how the dielectric constant provides insights into the efficiency of a dielectric material in modifying the electric field within a capacitor.
Bound Charge
In dielectric materials, the concept of bound charge is important to understand the internal electric effects due to polarization. When an external electric field is applied, the positive and negative charges in the dielectric material shift slightly in opposite directions. This creates a surface charge on the dielectric, known as the bound charge.

The bound surface charge density (\textbf{蟽\(_b\)}) can be calculated using the following formula:

\textbf{蟽\(_b\)} = Q \times \bigg(1 - \frac{1}{\textbf{魏}}\bigg)

Inserting the given values:

\textbf{蟽\(_b\)} = 8.9 \times 10^{-7} \times \bigg(1 - \frac{1}{71788.6}\bigg) \textbf{鈮 8.9 \times 10^{-7} C/m虏}

The bound charge on each surface (\textbf{Q\(_b\)}) is then given by multiplying the bound surface charge density with the area (\textbf{A}) of the plates:

\textbf{Q\(_b\)} = \textbf{蟽\(_b\)} \times \textbf{A} = 8.9 \times 10^{-7} \times 0.01 \textbf{鈮 8.9 \times 10^{-9} C}

Bound charge reflects the internal electric response of the dielectric and contributes significantly to the effectiveness of the dielectric in a capacitor system.

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Most popular questions from this chapter

A controller on an electronics arcade games consists of a variable resistor connected across the plates of a \(0.220 \mu \mathrm{F}\) capacitor. The capacitor is charged to \(5.00 \mathrm{~V}\), then discharged through the resistor. The time for the potential difference across the plates to decrease to \(0.800 \mathrm{~V}\) is measured by a clock inside the game. If the range of discharge times that can be handled effectively is from \(10.0 \mu\) s to \(6.00 \mathrm{~ms}\), what should be the resistance range of the resistor?

A \(1.0 \mu \mathrm{F}\) capacitor with an initial stored energy of \(0.50 \mathrm{~J}\) is discharged through a \(1.0 \mathrm{M} \Omega\) resistor. (a) What is the initial amount of excess charge on the capacitor plates? (b) What is the current through the resistor when the discharge starts? (c) Determine \(\Delta V_{C}\), the potential difference across the capacitor, and \(\Delta V_{R}\), the potential difference across the resistor, as functions of time. (d) Express the production rate of thermal energy in the resistor as a function of time.

A capacitor with initial excess charge of amount \(\left|q_{0}\right|\) is discharged through a resistor. In terms of the time constant \(\tau\), how long is required for the capacitor to lose (a) the first one-third of its charge and (b) two-thirds of its charge?

(a) Show that the plates of a parallel-plate capacitor attract each other with a force of magnitude given by \(F=q^{2} / 2 \varepsilon_{0} A .\) Do so by calculating the work needed to increase the plate separation from \(x\) to \(x+d x\), with the excess charge \(|q|\) remaining constant. (b) Next show that the magnitude of the force per unit area (the electrostatic stress) acting on either capacitor plate is given by \(\frac{1}{2} \varepsilon_{0} E^{2}\). (Actually, this is the force per unit area on any conductor of any shape with an electric field \(\vec{E}\) at its surface.)

\(100 \mathrm{pF}\) capacitor is charged to a potential difference of \(50 \mathrm{~V}\), and the charging battery is disconnected. The capacitor is then connected in parallel with a second (initially uncharged) \(\quad\) capacitor. If the potential difference across the first capacitor drops to \(35 \mathrm{~V}\) what is the capacitance of this second capacitor?

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