/*! 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 6 How far from a \(-7.20-\mu \math... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

How far from a \(-7.20-\mu \mathrm{C}\) point charge must \(\mathrm{a}+2.30-\mu \mathrm{C}\) point charge be placed for the electric potential energy \(U\) of the pair of charges to be \(-0.400 \mathrm{J}\) ? (Take \(U\) to be zero when the charges have infinite separation.)

Short Answer

Expert verified
The separation distance must be approximately 0.37 meters.

Step by step solution

01

Understand the Electric Potential Energy Formula

The formula for electric potential energy between two point charges is given by \( U = \frac{k \cdot q_1 \cdot q_2}{r} \), where \( k \) is Coulomb's constant \( (8.988 \times 10^9 \frac{N \cdot m^2}{C^2}) \), \( q_1 \) and \( q_2 \) are the charges, and \( r \) is the distance between the charges.
02

Identify Known Quantities

The charges are given as \( q_1 = -7.20 \times 10^{-6} \mathrm{C} \) and \( q_2 = 2.30 \times 10^{-6} \mathrm{C} \). The electric potential energy \( U \) is \(-0.400 \mathrm{J} \). Coulomb's constant is \( k = 8.98 \times 10^9 \frac{N \cdot m^2}{C^2} \).
03

Substitute Known Values into the Formula

Using the formula \( U = \frac{k \cdot q_1 \cdot q_2}{r} \), substitute the known values: \(-0.400 = \frac{(8.98 \times 10^9) \cdot (-7.20 \times 10^{-6}) \cdot (2.30 \times 10^{-6})}{r} \).
04

Solve for the Distance \( r \)

Rearrange the equation to solve for \( r \): \[ r = \frac{(8.98 \times 10^9) \times (-7.20 \times 10^{-6}) \times (2.30 \times 10^{-6})}{-0.400} \]. Calculate the value to find \( r \).
05

Calculate Numerically

Perform the arithmetic: the numerator is \( (8.98 \times 10^9) \times (-7.20 \times 10^{-6}) \times (2.30 \times 10^{-6}) = -0.1484736 \), and dividing by \(-0.400\) gives the distance \( r \). So, \[ r = \frac{0.1484736}{0.400} = 0.37118 \] meters.

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.

Point Charges
Point charges are a fundamental concept in physics, representing charged objects that are so small in size that their shape doesn't affect the calculations of electric forces and fields. These charges are assumed to be concentrated at a single point in space. This simplification allows us to easily apply mathematical formulas to understand and predict how these charges interact with each other.
Understanding point charges is essential because they are the building blocks for analyzing electric fields and potential energy in systems. In most introductory physics problems, such as calculating the electric potential energy between two charges, we use point charges to simplify complex interactions.
  • Point charges help us focus on the electric properties, leaving out unnecessary details.
  • They enable straightforward use of formulas like Coulomb's law.
In the original exercise, you are dealing with two point charges: one with the charge of (-7.20 ext{ μC}) and the other with (+2.30 ext{ μC}) place emphasis on the fundamental nature of point charges. These values make it easy to calculate interactions using theoretical formulas.
Coulomb's Law
Coulomb's Law provides a powerful tool for determining the force or energy between two point charges. It is expressed mathematically to quantify the electric interaction in terms of forces and energy between charges and their separation.
The formula derived from this law is expressed as:\[ F = \frac{k \cdot |q_1| \cdot |q_2|}{r^2} \]where:
  • \( F \) is the magnitude of the electric force between the charges.
  • \( k \) is Coulomb's constant, approximately \( 8.988 \times 10^9 \frac{N \cdot m^2}{C^2} \).
  • \( q_1 \) and \( q_2 \) are the magnitudes of the two charges.
  • \( r \) is the distance between the two charges.
In the context of calculating potential energy, Coulomb's Law helps us understand how these forces translate into stored energy when charges interact. The energy is tied directly to their positions in space.
The original exercise utilizes this law's principles to find the distance where a specific potential energy is achieved, showcasing how force interaction concepts are inherently linked with energy calculation.
Distance Calculation
Distance calculation is an important aspect of problems involving point charges and potential energy. The ability to determine how far apart two charges must be to achieve a certain electrical effect is key in physics.
In the exercise, we solve for the distance \( r \) using the formula for electric potential energy:\[ U = \frac{k \cdot q_1 \cdot q_2}{r} \]Let's break down how the distance is found:
  1. Identify the given values for both charges and the desired potential energy.
  2. Rearrange the formula to solve for \( r \):\[ r = \frac{k \cdot q_1 \cdot q_2}{U} \]
  3. Substitute the known quantities into the equation and solve for \( r \).
Distance calculation in this context means solving a practical and mathematical problem to arrange the point charges to achieve specific energy conditions. Clear understanding of this calculation helps bridge the gap between theoretical concepts and real-world physics applications.
This method is not only essential in educational exercises but also in practical electrical engineering and physics applications.
Physics Problem Solving
Physics problem solving involves a step-by-step approach to understanding problems and deriving solutions based on physics principles. Whether you are new to physics or a seasoned learner, approaching problems methodically ensures clarity and comprehensibility.
Here's a simple strategy to tackle such problems:
  • **Understand the Problem**: Carefully read and interpret what is being asked. Identify what you know and what you need to find out.
  • **Identify the Clues**: Pinpoint key details such as formulas that might relate to the problem. In the task at hand, understanding key concepts like electric potential energy and Coulomb's law is crucial.
  • **Execute the Mathematics**: Plug values into formulas and solve. Remember to keep track of units and algebraic manipulations.
  • **Verify Results**: Check if your solution is reasonable in the context of the problem.
The original exercise on finding the right distance for a specific potential energy is a classic example of physics problem solving. Adopting structured methods like these enhances logical thinking and boosts confidence in tackling physics challenges.

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

Energy of the Nucleus. How much work is needed to assemble an atomic nucleus containing three protons (such as Be) if we model it as an equilateral triangle of side \(2.00 \times 10^{-13} \mathrm{m}\) with a proton at each vertex? Assume the protons started from very far away.

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.

(a) An electron is to be accelerated from \(3.00 \times 10^{6} \mathrm{m} / \mathrm{s}\) to \(8.00 \times 10^{6} \mathrm{m} / \mathrm{s}\) . Through what potential difference must the electron pass to accomplish this? (b) Through what potential difference must the electron pass if it is to be slowed from \(8.00 \times 10^{6} \mathrm{m} / \mathrm{s}\) to a halt?

The electric field at the surface of a charged, solid, copper 0sphere with radius 0.200 \(\mathrm{m}\) is 3800 \(\mathrm{N} / \mathrm{C}\) , directed toward the center of the sphere. What is the potential at the center of the sphere, if we take the potential to be zero infinitely far from the sphere?

A very long wire carries a uniform linear charge density \(\lambda\) . Using a voltmeter to measure potential difference, you find that when one probe of the meter is placed 2.50 \(\mathrm{cm}\) from the wire and the other probe is 1.00 \(\mathrm{cm}\) farther from the wire, the meter reads 575 \(\mathrm{V}\) . (a) What is \(\lambda 2\) (b) If you now place one probe at 3.50 \(\mathrm{cm}\) from the wire and the other probe 1.00 \(\mathrm{cm}\) farther away, will the voltmeter read 575 \(\mathrm{V} ?\) If not, will it read more or less than 575 \(\mathrm{V}\) ? Why? (c) If you place both probes 3.50 \(\mathrm{cm}\) from the wire but 17.0 \(\mathrm{cm}\) from each other, what will the voltmeter read?

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.