The electric field strength is 20,000N/Cinside a parallel-plate capacitor with a 1.0mm spacing. An electron is released from rest at the negative plate. What is the electron’s speed when it reaches the positive plate?

Short Answer

Expert verified

Electron speed is2.7106m/swhen it reaches the positive speed.

Step by step solution

01

Step :1 Introduction 

According to the law of conservation of momentum, all energy generated from deformation in an electrostatic potential is transitioned to angular momentum. This enables us to compose.

qΔV=mv22

where can we discover out what the speed will be

v=2qΔVm

02

Step :2  Theoretical difference 

Everything we really need to do is determine to see what the alternative potential is. Realizing that the electromagnetic field within a resistor has a magnitude of

E=ΔVd

The theoretical discrepancy can be identified as follows:

ΔV=Ed

03

Step :3  Substitution 

By swapping this, we can reach at our finalized performance articulation:

v=2qEdm

Consequently, when we apply proportionally, we get

v=21.61019210411039.11031=2.7106m/s

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 Vaia!

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 -10.0 nC point charge and a +20.0 nC point charge are 15.0 cm apart on the x-axis. a. What is the electric potential at the point on the x-axis where the electric field is zero? b. What is the magnitude of the electric field at the point on the x-axis, between the charges, where the electric potential is zero?

x=-1.0cmA-2.0nCcharge and a +2.0nCcharge are located on the x-axis at and x=+1.0cm, respectively.

a. Other than at infinity, is there a position or positions on the x-axis where the electric field is zero? If so, where?

b. Other than at infinity, at what position or positions on the x-axis is the electric potential zero?

c. Sketch graphs of the electric field strength and the electric potential along the x-axis.

In the form of radioactive decay known as alpha decay, an unstable nucleus emits a helium-atom nucleus, which is called an alpha particle. An alpha particle contains two protons and two neutrons, thus having massm=4uand charge q=2e. Suppose a uranium nucleus with 92protons decays into thorium, with 90protons, and an alpha particle. The alpha particle is initially at rest at the surface of the thorium nucleus, which is 15fmin diameter. What is the speed of the alpha particle when it is detected in the laboratory? Assume the thorium nucleus remains at rest

Proton moves along the x-axis where some arrangement of charges has produced the potential vx=v0sin(2π/λ), where and l = 1.0 mm.

a. What minimum speed must the proton have at x = 0 to move down the axis without being reflected?

b. What is the maximum speed reached by a proton that at x = 0

Living cells “pump” singly ionized sodium ions, Na+, from the inside of the cell to the outside to maintain a membrane potential ∆Vmembrane = Vin - Vout = - 70 mV. It is called pumping because work must be done to move a positive ion from the negative inside of the cell to the positive outside, and it must go on continuously because sodium ions “leak” back through the cell wall by diffusion. a. How much work must be done to move one sodium ion from the inside of the cell to the outside? b. At rest, the human body uses energy at the rate of approximately 100 W to maintain basic metabolic functions. It has been estimated that 20% of this energy is used to operate the sodium pumps of the body. Estimate—to one significant figure—the number of sodium ions pumped per second.

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.

Sign-up for free