/*! 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 21 A very small sphere with positiv... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 3: Problem 21

A very small sphere with positive charge \(q=+8.00 \mu \mathrm{C}\) is released from rest at a point \(1.50 \mathrm{~cm}\) from a very long line of uniform linear charge density \(\lambda=+3.00 \mu \mathrm{C} / \mathrm{m}\). What is the kinetic energy of the sphere when it is \(4.50 \mathrm{~cm}\) from the line of charge if the only force on it is the force exerted by the line of charge?

Short Answer

Expert verified
The kinetic energy of the sphere when it is 4.50 cm from the line of charge is equal to the change in electric potential energy as it moved from its initial position.

Step by step solution

01

Calculate Electric Potential at Initial and Final Positions

The electric potential due to an infinite line of charge at a distance \( r \) from it is given by the expression:\[ V(r) = -\frac{2k_e \lambda}{ r } \].First calculate the potential at the initial distance \( r_1 = 1.50 \, \text{cm} = 0.015 \, \text{m} \):\[ V_i = -\frac{2 \cdot 8.99 \times 10^9 \, \frac{\text{Nm}^2}{\text{C}^2} \cdot 3.00 \times 10^{-6} \, \frac{\text{C}}{\text{m}}}{0.015 \, \text{m}} \].Now calculate the potential at the final distance \( r_2 = 4.50 \, \text{cm} = 0.045 \, \text{m} \):\[ V_f = -\frac{2 \cdot 8.99 \times 10^9 \, \frac{\text{Nm}^2}{\text{C}^2} \cdot 3.00 \times 10^{-6} \, \frac{\text{C}}{\text{m}}}{0.045 \, \text{m}} \].
02

Determine the Change in Electric Potential Energy

The change in electric potential energy for the charge \( q \) moving in an electric field is given by:\[ \Delta U = q(V_f - V_i) \].Substitute the values of potential found in the first step to find \( \Delta U \):\[ \Delta U = 8.00 \times 10^{-6} \, \text{C} \times (V_f - V_i) \].
03

Calculate the Kinetic Energy of the Sphere

By conservation of energy, the change in potential energy \( \Delta U \) is equal to the change in kinetic energy \( \Delta K \). Since the sphere starts from rest, its initial kinetic energy is zero:\[ \Delta K = \Delta U = q(V_f - V_i) \].Evaluate this to find the kinetic energy of the sphere at the final position.

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.

Electric Potential
Electric potential, also known as voltage, describes the potential energy per unit charge at a specific point in an electric field. It's like the pressure in a water system that can push water through a pipe. The electric potential due to a line of charge decreases as you move further away from it because the influence of the charge diminishes.
This potential is calculated using the formula:
  • \[ V(r) = -\frac{2k_e \lambda}{r} \]
where:
  • \( V(r) \) is the electric potential at a distance \( r \)
  • \( k_e \) is the Coulomb's constant \( 8.99 \times 10^9 \, \text{Nm}^2/\text{C}^2 \)
  • \( \lambda \) is the linear charge density of the line of charge
Using this formula, you can find the potential at any point around a line of charge. This is crucial because electric potential affects the movement of charges, similar to how gravity influences the motion of objects.
Linear Charge Density
Linear charge density represents how much charge exists along a certain length of a line, and it is noted as \( \lambda \). This value helps us understand the distribution of charge along objects like wires or rods.
It is measured in Coulombs per meter (C/m). For example, a linear charge density of \( +3.00 \, \mu \text{C/m} \) means there are 3.00 microcoulombs of charge for each meter.
Understanding linear charge density is essential when dealing with electric fields around objects because it can significantly influence the strength and direction of the field.
Conservation of Energy
The conservation of energy principle is a foundational concept in physics stating that energy cannot be created or destroyed, only transformed. In the context of electric charges, this principle helps us understand how potential energy is converted into kinetic energy or vice versa.
For a charge moving in an electric field, as in this exercise, the initial potential energy transforms into kinetic energy as the charge accelerates due to the field.
Mathematically, we can express this as:
  • \[ \Delta U = \Delta K \]
where:
  • \( \Delta U \) is the change in electric potential energy
  • \( \Delta K \) is the change in kinetic energy
This ensures that the total energy of the system remains constant, and it allows us to calculate how fast a charge will move when subjected to different electric potentials.
Electric Potential Energy
Electric potential energy is the energy a charge possesses because of its position in an electric field. It's the energy stored due to the charge's electrical interaction with other charges.
In this exercise, we evaluate how this energy changes as the charged sphere moves from one point to another around a line of charge.
The change in electric potential energy \( \Delta U \) as a charge moves is calculated using the expression:
  • \[ \Delta U = q(V_f - V_i) \]
where:
  • \( q \) is the charge
  • \( V_f \) is the final electric potential
  • \( V_i \) is the initial electric potential
This formula reflects how much potential energy the charge gains or loses as it moves within the electric field, leading to changes in its kinetic energy.

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 point charge \(q_{1}\) is held stationary at the origin. A second charge \(q_{2}\) is placed at point \(a\), and the electric potential energy of the pair of charges is \(+5.4 \times 10^{-8} \mathrm{~J}\). When the second charge is moved to point \(b\), the electric force on the charge does \(-1.9 \times 10^{-8} \mathrm{~J}\) of work. What is the electric potential energy of the pair of charges when the second charge is at point \(b\) ?

Charge \(Q=5.00 \mu \mathrm{C}\) is distributed uniformly over the volume of an insulating sphere that has radius \(R=12.0 \mathrm{~cm}\). A small sphere with charge \(q=+3.00 \mu \mathrm{C}\) and mass \(6.00 \times 10^{-5} \mathrm{~kg}\) is projected toward the center of the large sphere from an initial large distance. The large sphere is held at a fixed position and the small sphere can be treated as a point charge. What minimum speed must the small sphere have in order to come within \(8.00 \mathrm{~cm}\) of the surface of the large sphere?

In a certain region, a charge distribution exists that is spherically symmetric but nonuniform. That is, the volume charge density \(\rho(r)\) depends on the distance \(r\) from the center of the distribution but not on the spherical polar angles \(\theta\) and \(\phi .\) The electric potential \(V(r)\) due to this charge distribution is $$ V(r)=\left\\{\begin{array}{l} \frac{\rho_{0} a^{2}}{18 \epsilon_{0}}\left[1-3\left(\frac{r}{a}\right)^{2}+2\left(\frac{r}{a}\right)^{3}\right] \text { for } r \leq a \\ 0 \quad \text { for } r \geq a \end{array}\right. $$ where \(\rho_{0}\) is a constant having units of \(\mathrm{C} / \mathrm{m}^{3}\) and \(a\) is a constant having units of meters. (a) Derive expressions for \(\overrightarrow{\boldsymbol{E}}\) for the regions \(r \leq a\) and \(r \geq\) a. [Hint: Use Eq. (23.23).] Explain why \(\vec{E}\) has only a radial component. (b) Derive an expression for \(\rho(r)\) in each of the two regions \(r \leq a\) and \(r \geq\) a. [Hint: Use Gauss's law for two spherical shells, one of radius \(r\) and the other of radius \(r+d r\). The charge contained in the infinitesimal spherical shell of radius \(d r\) is \(\left.d q=4 \pi r^{2} \rho(r) d r .\right]\) (c) Show that the net charge contained in the volume of a sphere of radius greater than or equal to \(a\) is zero. [Hint: Integrate the expressions derived in part (b) for \(\rho(r)\) over a spherical volume of radius greater than or equal to \(a\).] Is this result consistent with the electric field for \(r>a\) that you calculated in part (a)?

* A metal sphere with radius \(R_{1}\) has a charge \(Q_{1} .\) Take the electric potential to be zero at an infinite distance from the sphere. (a) What are the electric field and electric potential at the surface of the sphere? This sphere is now connected by a long, thin conducting wire to another sphere of radius \(R_{2}\) that is several meters from the first sphere. Before the connection is made, this second sphere is uncharged. After electrostatic equilibrium has been reached, what are (b) the total charge on each sphere; (c) the electric potential at the surface of each sphere; (d) the electric field at the surface of each sphere? Assume that the amount of charge on the wire is much less than the charge on each sphere.

* Three point charges, which initially are infinitely far apart, are placed at the corners of an equilateral triangle with sides \(d\). Two of the point charges are identical and have charge \(q\). If zero net work is required to place the three charges at the corners of the triangle, what must the value of the third charge be?
See all solutions

Recommended explanations on Physics Textbooks

Magnetism

Read Explanation

Wave Optics

Read Explanation

Work Energy and Power

Read Explanation

Turning Points in Physics

Read Explanation

Force

Read Explanation

Radiation

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.