/*! 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 79 An electron is released from res... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

An electron is released from rest on the axis of an electric dipole that has charge \(e\) and charge separation \(d=20 \mathrm{pm}\) and that is fixed in place. The release point is on the positive side of the dipole, at distance \(7.0 d\) from the dipole center. What is the electron's speed when it reaches a point \(5.0 d\) from the dipole center?

Short Answer

Expert verified
The electron's speed is \( v = \sqrt{\frac{2e(V_i - V_f)}{m}} \).

Step by step solution

01

Understand the Problem

We need to find the speed of an electron as it moves under the influence of an electric dipole. The electron starts from rest at distance \(7.0 d\) and moves to a distance \(5.0 d\) from the center of the dipole.
02

Electric Potential of a Dipole

The electric potential \(V\) at a distance \(r\) from the center on the axis of a dipole is given by: \[ V = \frac{k \cdot p}{r^2} \]where \(k\) is Coulomb's constant and \(p = e \cdot d\) is the dipole moment.
03

Calculate Initial and Final Potentials

Calculate the initial potential \(V_i\) at \(7.0 d\): \[ V_i = \frac{k \cdot p}{(7.0d)^2} \]Calculate the final potential \(V_f\) at \(5.0 d\): \[ V_f = \frac{k \cdot p}{(5.0d)^2} \]
04

Calculate the Potential Energy Change

The change in potential energy \(\Delta U\) is given by:\[ \Delta U = e(V_f - V_i) \]
05

Use Conservation of Energy

By conservation of energy, the change in kinetic energy \(\Delta K\) is equal to negative the change in potential energy:\[ \Delta K = -\Delta U \]Since the electron starts from rest, initial kinetic energy is zero, and thus\[ \frac{1}{2}mv^2 = -e(V_f - V_i) \]
06

Solve for Electron's Speed

Solving for \(v\), the speed of the electron, gives:\[ v = \sqrt{\frac{2e(V_i - V_f)}{m}} \]where \(m\) is the mass of the electron.

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.

Electric Potential
Electric potential is a crucial concept when studying electric dipoles, as it describes the potential energy per unit charge at a point in an electric field. For a dipole, this potential is due to the two separated charges, one positive and one negative, that create an electric field between and around them. The electric potential at a point a distance \( r \) from the middle of the dipole on its axis is given by the formula:
\[ V = \frac{k \cdot p}{r^2} \]
where:
  • \( k \) is the Coulomb's constant (approximately \( 8.99 \times 10^9 \text{ N m}^2/\text{C}^2 \)).
  • \( p \), the dipole moment, is the product of the charge \( e \) and the separation distance \( d \) between the charges.
Understanding electric potential helps us determine the work done on a charge, like the electron in the problem, as it moves in the electric field created by the dipole. As we move from one point to another in this field, the electric potential changes, impacting the potential energy of a charge placed in that field.
Conservation of Energy
Conservation of energy is a fundamental principle in physics stating that energy cannot be created or destroyed in an isolated system, only transformed from one form to another. In the context of an electron moving in the field of a dipole, as given in the problem, this principle allows us to relate changes in electric potential energy to changes in kinetic energy.
The electron starts at a certain electric potential, and as it moves, its potential energy changes according to the electric potential difference between its starting and ending positions:
\[ \Delta U = e(V_f - V_i) \]
Since the electron starts from rest, its kinetic energy initially is zero. Thus, the gain in kinetic energy \( \Delta K \) as it moves is equal to the decrease in its potential energy:
  • \( \Delta K = -\Delta U \).
This means:
\[ \frac{1}{2}mv^2 = -e(V_f - V_i) \]
This energy transformation is key in determining the electron's final speed, which requires calculating the change in potential due to its movement within the electric dipole field.
Dipole Moment
The dipole moment is a vector quantity representing the strength and orientation of an electric dipole. It’s an essential parameter in understanding the behavior of dipoles in electric fields. The dipole moment, \( p \), is given by the product of the charge \( e \) and the separation distance \( d \) between the positive and negative charges of the dipole:
\[ p = e \cdot d \]
This value provides a measure of how strong the dipole is and influences the electric potential and field around it. As in our problem, where the dipole is fixed with a certain charge separation, we can use the dipole moment to calculate the electric potential at various points along its axis.
  • The larger the dipole moment, the stronger the electric field and potential created by the dipole.
  • Knowing the dipole moment allows us to predict the forces on charges within the field, such as our electron moving through this electric field.
Thus, the dipole moment is not just a measure; it's a bridge between the microscopic world of charged particles and the macroscopic effects they induce in electric fields.

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 charge of \(1.50 \times 10^{-8} \mathrm{C}\) lies on an isolated metal sphere of radius \(16.0 \mathrm{~cm}\). With \(V=0\) at infinity, what is the electric potential at points on the sphere's surface?

The magnitude \(E\) of an electric field depends on the radial distance \(r\) according to \(E=A / r^{4}\), where \(A\) is a constant with the unit volt-cubic meter. As a multiple of \(\bar{A}\), what is the magnitude of the electric potential difference between \(r=2.00 \mathrm{~m}\) and \(r=3.00 \mathrm{~m} ?\)

An infinite nonconducting sheet has a surface charge density \(\sigma=+5.80 \mathrm{pC} / \mathrm{m}^{2} .\) (a) How much work is done by the electric field due to the sheet if a particle of charge \(q=+1.60 \times 10^{-19} \mathrm{C}\) is moved from the sheet to a point \(P\) at distance \(d=3.56 \mathrm{~cm}\) from the sheet? (b) If the electric potential \(V\) is defined to be zero on the sheet, what is \(V\) at \(P ?\)

The electric potential at points in an \(x y\) plane is given by \(V=\left(2.0 \mathrm{~V} / \mathrm{m}^{2}\right) x^{2}-\left(3.0 \mathrm{~V} / \mathrm{m}^{2}\right) y^{2} .\) In unit-vector notation, what is the electric field at the point \((3.0 \mathrm{~m}, 2.0 \mathrm{~m})\) ?

Two charged, parallel, flat conducting surfaces are spaced \(d=\) \(1.00 \mathrm{~cm}\) apart and produce a potential difference \(\Delta V=625 \mathrm{~V}\) between them. An electron is projected from one surface directly toward the second. What is the initial speed of the electron if it stops inst at the second surface?

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.