/*! 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 8 Three equal \(1.20-\mu \mathrm{C... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 23: Problem 8

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.)

Short Answer

Expert verified
The potential energy of the system is approximately \( 7.77 \times 10^{-2} \ \mathrm{J} \).

Step by step solution

01

Understand the concept

The potential energy of a system of point charges is calculated based on the pairwise interactions between them. For each pair of charges, we use the formula for potential energy: \[ U = k \frac{q_1 q_2}{r} \]where \( k \) is Coulomb's constant, \( q_1 \) and \( q_2 \) are the magnitudes of the charges, and \( r \) is the distance between the charges.
02

Write down the knowns

We know the following values:- Each charge, \( q = 1.20 \times 10^{-6} \ \mathrm{C} \)- Distance between each pair, \( r = 0.500 \ \mathrm{m} \)- Coulomb's constant, \( k = 8.99 \times 10^9 \ \mathrm{N} \cdot \mathrm{m}^2/\mathrm{C}^2 \)
03

Calculate the potential energy for one pair

Using the potential energy formula for one pair:\[ U_{pair} = k \frac{(1.20 \times 10^{-6})(1.20 \times 10^{-6})}{0.500} \] Compute: \[ U_{pair} = (8.99 \times 10^9) \frac{1.44 \times 10^{-12}}{0.500} \approx 2.59 \times 10^{-2} \ \mathrm{J} \]
04

Determine the total potential energy of the system

Since the triangle is equilateral, we have three pairs of equal charges. Therefore, the total potential energy \( U \) is three times the potential energy of one pair:\[ U = 3 \times U_{pair} = 3 \times 2.59 \times 10^{-2} \] Calculate:\[ U \approx 7.77 \times 10^{-2} \ \mathrm{J} \]

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.

Coulomb's Law
Coulomb's Law is a fundamental principle in electrostatics that describes the force between two point charges. This law tells us how charges attract or repel each other with a force. The magnitude of this force is directly proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance between them.
This can be expressed as:\[ F = k \frac{|q_1 q_2|}{r^2} \]where:
  • \( F \) is the magnitude of the force between the charges.
  • \( k \) is Coulomb's constant \((8.99 \times 10^9 \ \mathrm{N} \cdot \mathrm{m}^2/\mathrm{C}^2)\).
  • \( q_1 \) and \( q_2 \) are the magnitudes of the two charges.
  • \( r \) is the distance between the two charges.
Coulomb's Law helps us understand interactions between electrical charges. In the problem, it also forms the cornerstone to calculate potential energy between charges. By knowing how much work it takes to bring charges from infinity to a specific configuration, we can then find potential energy using a modified version of the formula by removing the square from \( r \).
Point Charges
Point charges are idealized charges located at a single point in space. This concept simplifies calculations in electrostatics, as we can ignore any physical size or shape. In real life, charges are often spread out over a volume; however, considering them as point charges lets us focus solely on their effects without complex geometry.
Each of the three charges in our problem is treated as a point charge, with a charge of \(1.20 \mu \mathrm{C} (1.20 \times 10^{-6} \, \mathrm{C})\). By using point charges, we apply Coulomb's Law to evaluate the potential energy of their configuration easily.
Point charges also help us ignore mutual distances. No matter where each point charge is within our equilateral triangle, the interactions are simply calculated using their distances, ensuring the method holds true even when dealing with complex arrangements.
Equilateral Triangle
An equilateral triangle is a special type of triangle where all three sides are equal in length and all three angles are equal, each measuring 60 degrees. This symmetry is helpful in calculating the potential energy in our setup as it suggests that the interactions between each pair of charges will be identical.
When considering electric potential energy in an equilateral triangle, you can assume that each charge sits at a vertex and that each side, which is the distance between charges, is equal. In our case, each side length is given as 0.500 meters.
This symmetry means that for each pair of charges, the calculation of potential energy using \( U = k \frac{q_1 q_2}{r} \) will yield the same result, simplifying our task to multiplying the potential energy found for one pair by three, since there are three identical pairs of charges in the triangle. This clarity significantly reduces the complexity of our analysis, allowing us to find the total potential energy with simple arithmetic after calculating a single pair's contribution.

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

In the Bohr model of the hydrogen atom, a single electron revolves around a single proton in a circle of radius \(r .\) Assume that the proton remains at rest. (a) By equating the electric force to the electron mass times its acceleration, derive an expression for the electron's speed. (b) Obtain an expression for the electron's kinetic energy, and show that its magnitude is just half that of the electric potential energy. (c) Obtain an expression for the total energy, and evaluate it using \(r=5.29 \times\) \(10^{-11} \mathrm{m} .\) Give your numerical result in joules and in electron volts.

The electric potential \(V\) in a region of space is given by $$V(x, y, z)=A\left(x^{2}-3 y^{2}+z^{2}\right)$$ where \(A\) is a constant. (a) Derive an expression for the electric field \(\vec{E}\) at any point in this region. (b) The work done by the field when a \(1.50-\mu \mathrm{C} \quad\) test charge moves from the point \((x, y, z)=\) \((0,0,0.250 \mathrm{m})\) to the origin is measured to be \(6.00 \times 10^{-5} \mathrm{J}\). Determine \(A .\) (c) Determine the electric field at the point \((0,0,\) 0.250 \(\mathrm{m} ) .\) (d) Show that in every plane parallel to the \(x z\) -plane the equipotential contours are circles. (e) What is the radius of the equipotential contour corresponding to \(V=1280 \mathrm{V}\) and \(y=2.00 \mathrm{m} ?\)

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 \(x>\) a? Explain why this result is obtained.

Two charges of equal magnitude \(Q\) are held a distance \(d\) apart. Consider only points on the line passing through both charges. (a) If the two charges have the same sign, find the location of all points (if there are any) at which (i) the potential (relative to infinity is zero (is the electric field zero at these points? (ii) the electric field is zero (is the potential zero at these points?). (b) Repeat part (a) for two charges having opposite signs.

A uniform electric field has magnitude \(E\) and is directed in the negative \(x\) -direction. The potential difference between point \(a\) (at \(x=0.60 \mathrm{m} )\) and point \(b\) (at \(x=0.90 \mathrm{m} )\) is 240 \(\mathrm{V}\) . (a) Which point, \(a\) or \(b,\) is at the higher potential? (b) Calculate the value of \(E.\) (c) A negative point charge \(q=-0.200 \mu C\) is moved from \(b\) to \(a\) . Calculate the work done on the point charge by the electric field.
See all solutions

Recommended explanations on Physics Textbooks

Energy Physics

Read Explanation

Further Mechanics and Thermal Physics

Read Explanation

Electrostatics

Read Explanation

Space Physics

Read Explanation

Famous Physicists

Read Explanation

Modern 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.