/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 34 We have a parallel-plate capacit... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

We have a parallel-plate capacitor, with each plate having a width \(W\) and a length \(L\). The plates are separated by air with a distance \(d\). Assume that \(L\) and \(W\) are both much larger than \(d\). The maximum voltage that can be applied is limited to \(V_{\max }=K d\), in which \(K\) is called the breakdown strength of the dielectric. Derive an expression for the maximum energy that can be stored in the capacitor in terms of \(K\) and the volume of the dielectric. If we want to store the maximum energy per unit volume, does it matter what values are chosen for \(L, W\), and \(d ?\) What parameters are important?

Short Answer

Expert verified
Energy per unit volume depends only on \(K\), not on \(L\), \(W\), or \(d\); thus, \(K\) is the key parameter.

Step by step solution

01

Understand the Problem

We have a parallel-plate capacitor with plates of dimensions length \(L\) and width \(W\), separated by air at distance \(d\). The maximum voltage is limited by the breakdown strength \(K\) of the dielectric material. We need to find the maximum energy stored in terms of \(K\) and the volume of the dielectric. Ultimately, we are to determine if \(L\), \(W\), and \(d\) affect the energy per unit volume.
02

Formula for Stored Energy

The energy \(U\) stored in a capacitor is given by the formula: \[ U = \frac{1}{2} C V^2 \] where \(C\) is the capacitance and \(V\) is the voltage across the capacitor. We use this formula to express energy in terms of our given parameters.
03

Find Capacitance Formula

For a parallel-plate capacitor, the capacitance \(C\) is given by: \[ C = \varepsilon_0 \frac{A}{d} \]where \(\varepsilon_0\) is the permittivity of free space and \(A = L \times W\) is the area of the plates.
04

Substitute for Capacitance and Maximum Voltage

Plug the expression for capacitance and maximum voltage into the energy formula: \[ U = \frac{1}{2} \left(\varepsilon_0 \frac{LW}{d}\right) (Kd)^2 \]Simplifying, we find \[ U = \frac{1}{2} \varepsilon_0 L W d K^2 \].
05

Consider Volume of Dielectric

The volume of the dielectric (air, in this case) is given by: \[ V_{dielectric} = L \times W \times d \].Hence, we can express the energy in terms of the volume: \[ U = \frac{1}{2} \varepsilon_0 V_{dielectric} K^2 \].
06

Determine Energy per Unit Volume

The energy per unit volume \(u\) is given by:\[ u = \frac{U}{V_{dielectric}} = \frac{1}{2} \varepsilon_0 K^2 \] This expression shows that the energy per unit volume depends only on \(K\) and is independent of \(L\), \(W\), or \(d\).
07

Determine Important Parameters

From the expression for energy per unit volume, the crucial parameters are the breakdown strength \(K\) and the permittivity \(\varepsilon_0\). Choices of \(L\), \(W\), and \(d\) affect the total volume but not the energy density (energy per unit volume).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Parallel-Plate Capacitor
A parallel-plate capacitor consists of two large, flat plates arranged facing each other. These plates have dimensions defined by their length \(L\) and width \(W\). They are separated by a distance \(d\), which is significantly smaller compared to the length and width of the plates. This setup allows the plates to store electrical energy.In the space between the plates, a material called a dielectric can be placed, which in this case is air. A dielectric increases the capacitance of the capacitor by reducing the electric field for a given charge on the plates. When a voltage is applied across the capacitor, an electric field develops between the plates, storing energy within this field. The simplicity and effectiveness of this arrangement make it a staple in many electronic and physical applications.
Breakdown Strength
Breakdown strength is a critical parameter in defining the operating limits of a capacitor. It represents the maximum electric field that a dielectric material (like air, in this scenario) can withstand without its insulating properties breaking down and becoming conductive. This is the point where the dielectric material can no longer prevent the flow of current and potentially allows a short circuit between the capacitor plates.The value of breakdown strength, denoted as \(K\), determines the maximum electric field and, hence, the maximum voltage \(V_{\text{max}}=K d\) that can be safely applied to the capacitor. It's crucial to choose a dielectric material with a suitable breakdown strength to ensure reliable capacitor performance and avoid damage or failure.
Energy Density
Energy density in a capacitor refers to the amount of energy stored per unit volume. For our parallel-plate capacitor, it helps us understand how efficiently the space between the plates is utilized in terms of energy storage. The formula for the energy density \(u\) is given by:\[ u = \frac{U}{V_{\text{dielectric}}} = \frac{1}{2} \varepsilon_0 K^2 \]where \(U\) is the total energy stored, \(V_{\text{dielectric}}\) is the volume of the space between the plates, and \(\varepsilon_0\) is the permittivity of free space. From this equation, it's clear that the energy density is dependent on the breakdown strength \(K\) and is independent of the dimensions \(L\), \(W\), and \(d\) of the capacitor itself. This independence emphasizes the role of the dielectric material's properties over physical size for energy storage efficiency.
Capacitance Formula
Capacitance is a measure of a capacitor's ability to store charge per unit voltage. For a parallel-plate capacitor, the capacitance \(C\) is determined by the formula:\[ C = \varepsilon_0 \frac{A}{d} \]where \(A = L \times W\) is the area of the plates, and \(d\) is the separation distance between them.The capacitance increases with the increase in plate area and decreases with a wider separation between the plates. Therefore, larger areas enable more charge storage capacity, while smaller separation enhances the electric field strength for a given charge, thus increasing capacitance. This relationship reminds students that size and relative placement of plates influence a capacitor's performance but not the energy stored per unit volume when the breakdown strength \(K\) is considered.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A pair of mutually coupled inductances has \(L_{1}=2 \mathrm{H}, L_{2}=1 \mathrm{H}, i_{1}=2 \cos (1000 t) \mathrm{A}\), \(i_{2}=0\), and \(v_{2}=2000 \sin (1000 t)\) V. (The arguments of the sine and cosine functions are in radians.) Find \(v_{1}(t)\) and the magnitude of the mutual inductance.

The maximum value obtained by connecting 4 identical inductances in series is \(16 \mathrm{H}\). What is the value of each inductor? What is the minimum value achieved when they are connected in parallel?

Suppose that a parallel-plate capacitor has a dielectric that breaks down if the electric field exceeds \(K \mathrm{~V} / \mathrm{m}\). Thus, the maximum voltage rating of the capacitor is \(V_{\max }=K d\), where \(d\) is the thickness of the dielectric. In working Problem P3.34, we find that the maximum energy that can be stored before breakdown is \(w_{\max }=1 / 2 \epsilon_{r} \epsilon_{0} K^{2}(\mathrm{Vol})\), in which Vol is the volume of the dielectric. Air has approximately \(K=32 \times 10^{5} \mathrm{~V} / \mathrm{m}\) and \(\epsilon_{r}=1\). Find the minimum volume of air (as a dielectric in a parallel-plate capacitor) needed to store the energy content of one U.S. gallon of gasoline, which is approximately \(132 \mathrm{MJ}\). What thickness should the air dielectric have if we want the voltage for maximum energy storage to be \(1000 \mathrm{~V}\) ?

A parallel-plate capacitor like that shown in Figure P3.36 has a capacitance of 2500 \(\mathrm{pF}\) when the tank is full so the plates are totally immersed in the insulating liquid. (The dielectric constant of the fluid and the plate dimensions are different for this problem than for Problem P3.36.) The capacitance is \(100 \mathrm{pF}\) when the tank is empty and the space between the plates is filled with air. Suppose that the tank is full and the capacitance is charged to \(1000 \mathrm{~V}\). Then, the capacitance is open circuited so the charge on the plates cannot change, and the tank is drained. Compute the voltage after the tank is drained and the electrical energy stored in the capacitor before and after the tank is drained. With the plates open circuited, there is no electrical source for the extra energy. Where could it have come from?

An unknown inductance \(L\) in series with a 4-H inductor is connected in parallel with a \(10-\mathrm{H}\) inductor. The effective inductance is \(5-\mathrm{H}\). Find the value of \(L\).

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.