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Show that for a given dielectric material the maximum energy a parallel plate capacitor can store is directly proportional to the volume of dielectric (Volume \(=A \cdot d\) ). Note that the applied voltage is limited by the dielectric strength.

Short Answer

Expert verified
The energy \( U \) stored is directly proportional to the dielectric volume.

Step by step solution

01

Understand the Capacitor Energy Formula

The energy stored in a parallel plate 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.
02

Capacitance Formula for Dielectric Material

The capacitance of a parallel plate capacitor with a dielectric is \( C = \varepsilon \frac{A}{d} \), where \( \varepsilon \) is the permittivity of the dielectric material, \(A\) is the area of the plates, and \(d\) is the separation between the plates.
03

Voltage and Dielectric Strength Relation

The maximum voltage \( V_{max} \) applicable across the dielectric is limited by its dielectric strength, given by \( V_{max} = E_s \cdot d \), where \( E_s \) is the dielectric strength.
04

Substitute and Simplify Energy Formula

Substitute the expression for \(C\) and \(V_{max}\) into the energy formula: \( U = \frac{1}{2} \varepsilon \frac{A}{d} (E_s \cdot d)^2 \). Simplify to \( U = \frac{1}{2} \varepsilon A E_s^2 d \).
05

Relate Energy to Volume

Recognize that the volume \( V = A \cdot d \). Substitute into the energy equation to get: \( U = \frac{1}{2} \varepsilon E_s^2 V \).
06

Conclude Proportionality

From \( U = \frac{1}{2} \varepsilon E_s^2 V \), we conclude that the energy \( U \) stored in the capacitor is directly proportional to the volume \( V \) of the dielectric.

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Key Concepts

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

Dielectric Material
A dielectric material is an insulating substance placed between the two plates of a capacitor. It serves two main functions: increasing the capacitor's capacitance and providing electrical insulation.
Dielectric materials are characterized by their permittivity, denoted as \( \varepsilon \). The permittivity describes how much electric field is reduced inside the material compared to a vacuum. A higher permittivity increases a capacitor's ability to store charge.
These materials allow a capacitor to store more energy at the same voltage or operate effectively at a lower voltage.
A dielectric isn't just limited to solid materials like glass or ceramic; liquids and gases can also act as dielectrics depending on their properties.
When selecting a dielectric, it's crucial to consider not only the permittivity but also the dielectric strength.
Parallel Plate Capacitor
A parallel plate capacitor consists of two conductive plates separated by a gap, filled with a dielectric material. The plates are electrically charged, one positive and one negative, setting up an electric field across the dielectric.
The capacitance \( C \) of such a capacitor without a dielectric is given by \( C = \varepsilon_0 \frac{A}{d} \), where \( \varepsilon_0 \) is the permittivity of free space, \( A \) is the area of the plates, and \( d \) is the distance between them.
With the presence of a dielectric, the formula modifies to \( C = \varepsilon \frac{A}{d} \), incorporating the dielectric’s permittivity \( \varepsilon \).
Increasing the plate area \( A \) or using a material with higher permittivity boosts the capacitance, allowing more energy to be stored. However, reducing the plate separation \( d \) also increases capacitance but might lead to electrical breakdown.
Dielectric Strength
Dielectric strength is a critical property of dielectric materials. It refers to the maximum electric field a dielectric can withstand without breakdown. Breakdown occurs when the insulating properties fail, and the dielectric becomes conductive.
This property is essential to ensure the safe operation of capacitors. It sets a limit on the voltage applied across the dielectric, known as \( V_{max} \).
The maximum voltage for a capacitor can be expressed as \( V_{max} = E_s \cdot d \), where \( E_s \) is the dielectric strength and \( d \) is the distance between the plates.
Higher dielectric strength means the material can endure stronger fields, allowing capacitors to be charged to higher voltages.
  • Dielectric strength is expressed in volts per meter (V/m).
  • The choice of dielectric is crucial: better materials ensure that capacitors can operate safely under higher voltages.
This is important for applications requiring high-power processing or those in risk-prone environments.

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Most popular questions from this chapter

Two parallel conducting plates are separated by \(10.0 \mathrm{~cm}\), and one of them is taken to be at zero volts. (a) What is the electric field strength between them, if the potential \(8.00 \mathrm{~cm}\) from the zero volt plate (and \(2.00 \mathrm{~cm}\) from the other) is \(450 \mathrm{~V} ?\) (b) What is the voltage between the plates?

An electrostatic paint sprayer has a 0.200-m-diameter metal sphere at a potential of \(25.0 \mathrm{kV}\) that repels paint droplets onto a grounded object. (a) What charge is on the sphere? (b) What charge must a \(0.100-\mathrm{mg}\) drop of paint have to arrive at the object with a speed of \(10.0 \mathrm{~m} / \mathrm{s} ?\)

In one of the classic nuclear physics experiments at the beginning of the 20th century, an alpha particle was accelerated toward a gold nucleus, and its path was substantially deflected by the Coulomb interaction. If the energy of the doubly charged alpha nucleus was \(5.00 \mathrm{MeV}\), how close to the gold nucleus ( 79 protons) could it come before being deflected?

Membrane walls of living cells have surprisingly large electric fields across them due to separation of ions. (Membranes are discussed in some detail in Nerve Conduction-Electrocardiograms.) What is the voltage across an \(8.00 \mathrm{~nm}\) -thick membrane if the electric field strength across it is \(5.50 \mathrm{MV} / \mathrm{m}\) ? You may assume a uniform electric field.

If the potential due to a point charge is \(5.00 \times 10^{2} \mathrm{~V}\) at a distance of \(15.0 \mathrm{~m}\), what are the sign and magnitude of the charge?

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