/*! 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 A vacuum tube diode consists of ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 3: Problem 34

A vacuum tube diode consists of concentric cylindrical electrodes, the negative cathode and the positive anode. Because of the accumulation of charge near the cathode, the electric potential between the electrodes is not a linear function of the position, even with planar geometry, but is given by $$ V(x)=C x^{4 / 3} $$ where \(x\) is the distance from the cathode and \(C\) is a constant, characteristic of a particular diode and operating conditions. Obtain a formula for the electric field between the electrodes as a function of \(x\)

Short Answer

Expert verified
The electric field between the electrodes is \(E(x) = -\frac{4C}{3}x^{1/3}\).

Step by step solution

01

Understanding Electric Field from Potential

The electric field \(E\) is related to the electric potential \(V(x)\) through the negative gradient: \(E = -\frac{dV}{dx}\). This implies that by differentiating the potential function \(V(x)\), we can find the corresponding electric field as a function of position \(x\).
02

Differentiating the Potential Function

Given the potential function \(V(x) = Cx^{4/3}\), take its derivative with respect to \(x\):\[\frac{dV}{dx} = \frac{d}{dx}(Cx^{4/3}) = C \cdot \frac{4}{3}x^{1/3} = \frac{4C}{3}x^{1/3}.\]
03

Applying the Electric Field Formula

Using the formula for the electric field \(E = -\frac{dV}{dx}\), substitute the derivative:\[E(x) = -\left(\frac{4C}{3}x^{1/3}\right) = -\frac{4C}{3}x^{1/3}.\]

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 91影视!

Key Concepts

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

Vacuum Tube Diode
Vacuum tube diodes are fundamental components in the history of electronic devices. These diodes are typically composed of two main parts: the cathode and the anode. The cathode is the negatively charged electrode that emits electrons, while the anode is the positively charged electrode that collects them. This construction allows electrons to flow in only one direction from cathode to anode within the tube, essentially acting as a rectifier.

Originally, vacuum tube diodes were pivotal in radio, television, and early computing technologies. Their ability to convert alternating current (AC) signal input into a direct current (DC) output made them revolutionary before technologies like the transistor took over. These diodes operate in a vacuum, which minimizes the interaction between the electrons and other particles, thus allowing for smooth electron flow.
  • Cathode: Emits electrons into the vacuum.
  • Anode: Collects electrons, allowing current flow.
  • Function: Converts AC to DC by allowing electron flow in one direction.
The understanding of a vacuum tube diode's functioning helps us recognize the dynamics between the electrodes and how they influence the electric potential.
Electric Potential
Electric potential is a crucial concept to understand the behavior of electric fields in various settings, including vacuum tube diodes. Electric potential represents the potential energy per unit charge at a point in a field created by electric charges. It's a scalar quantity, usually denoted by \(V\), that indicates how much potential energy a unit charge would have at a specific location.

In the context of cylindrical electrodes in a vacuum tube diode, the potential difference between the cathode and anode is not linear, due to charge accumulation near the cathode. The exercise shows that the electric potential varies according to the equation \(V(x) = Cx^{4/3}\), where \(x\) is the distance from the cathode.
  • Formula: Described as \(V(x) = Cx^{4/3}\).
  • Non-linear variation: Due to charge buildup, the potential doesn't change linearly with distance.
  • Scalar nature: It doesn't have direction, only a magnitude.
Understanding electric potential helps predict how electrons will move between the electrodes and is key to determining the electric field.
Cylindrical Electrodes
Cylindrical electrodes play a significant role in the unique behavior observed in devices like vacuum tube diodes. These electrodes are coaxial, meaning they share a common central axis, which influences the electric field distribution between them. This setup often leads to a radial symmetry in the electric field.

The challenge with cylindrical electrodes in a vacuum tube diode is the non-linear relationship between distance and electric potential, caused by an accumulation of charge near the cathode. This can create a more complex electric potential profile as compared to linear setups. The formula \(V(x) = Cx^{4/3}\) reflects this complex interplay, showing how potential changes along the radius between these electrodes.
  • Coaxial arrangement: Ensures a uniform radial electric field distribution.
  • Non-linear potential: Results from cylindrical geometry and charge effects.
  • Electric field influence: The arrangement directly affects how the electric field is calculated.
Mastering the behavior of cylindrical electrodes helps in understanding the derived electric field formula and the physical processes within a vacuum tube diode.

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

* Three equal \(1.20-\mu \mathrm{C}\) point charges are placed at the corners of an equilateral triangle whose sides are \(0.500 \mathrm{~m}\) long. What is the potential energy of the system? (Take as zero the potential energy of the three charges when they are infinitely far apart.)

A disk with radius \(R\) has uniform surface charge density \(\sigma\). (a) By regarding the disk as a series of thin concentric rings, calculate the electric potential \(V\) at a point on the disk's axis a distance \(x\) from the center of the disk. Assume that the potential is zero at infinity. Calculate \(E_{n}=-\frac{d V}{d x}\)

A positive charge \(+q\) is located at the point \(x=0\), \(y=-a\), and a negative charge \(-q\) is located at the point \(x=0\), \(y=+a .\) (a) Derive an expression for the potential \(V\) at points on the \(y\)-axis as a function of the coordinate \(y\). Take \(V\) to be zero at an infinite distance from the charges. (b) Graph \(V\) at points on the \(y\)-axis as a function of \(y\) over the range from \(y=-4 a\) to \(y=+4 a\). (c) Show that for \(y>a\), the potential at a point on the positive \(y\)-axis is given by \(V=-\left(1 / 4 \pi \epsilon_{0}\right) 2 q a / y^{2}\). (d) What are the answers to parts (a) and (c) if the two charges are interchanged so that \(+q\) is at \(y=+a\) and \(-q\) is at \(y=-a\) ?

* Charge \(Q=+4.00 \mu \mathrm{C}\) is distributed uniformly over the volume of an insulating sphere that has radius \(R=5.00 \mathrm{~cm}\). What is the potential difference between the center of the sphere and the surface of the sphere?

A positive charge \(q\) is fixed at the point \(x=0, y=0\), and a negative charge \(-2 q\) is fixed at the point \(x=a, y=0\). (a) Show the positions of the charges in a diagram. (b) Derive an expression for the potential \(V\) at points on the \(x\)-axis as a function of the coordinate \(x\). Take \(V\) to be zero at an infinite distance from the charges. (c) At which positions on the \(x\)-axis is \(V=0 ?\) (d) Graph \(V\) at points on the \(x\)-axis as a function of \(x\) in the range from \(x=-2 a\) to \(x=+2 a .\) (e) What does the answer to part (b) become when \(x \gg a\) ? Explain why this result is obtained.
See all solutions

Recommended explanations on Physics Textbooks

Scientific Method Physics

Read Explanation

Fundamentals of Physics

Read Explanation

Turning Points in Physics

Read Explanation

Astrophysics

Read Explanation

Magnetism and Electromagnetic Induction

Read Explanation

Solid State Physics

Read Explanation
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.