/*! 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] | 91Ó°ÊÓ

91Ó°ÊÓ

Three equal \(1.20 \mu \mathrm{C}\) point charges are placed at the corners of an equilateral triangle with sides \(0.400 \mathrm{~m}\) 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 \( 8.10 Joules.\)

Step by step solution

01

Understand Concept

Potential energy of a system with two point charges is given by \(U = k \cdot \frac{q_1 \cdot q_2 }{r}\), where \(k\) is Coulomb's constant, \(q_1\) and \(q_2\) are the charges and \(r\) is the distance between the charges.
02

Apply Formula

Apply the formula for each pair of charges. Here \(k = 8.99 \times 10^9 N m^2/C^2\), \(q_1 = q_2 = 1.2 \times 10^{-6} C\) and \(r = 0.4m \). Thus, the potential energy between two charges is \(U = 8.99 \times 10^9 \cdot \frac{1.2 \times 10^{-6} \cdot 1.2 \times 10^{-6}}{0.4} = 2.6985 Joules.\)
03

Combine Potentials of Pairs

Since the system contains 3 identical pairs of charges, simply triple the potential calculated in step 2. So, the total potential energy of the system is \(3 \cdot 2.6985 = 8.0955 Joules.\)

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 fundamental to understanding electrostatics. Imagine each point charge as a tiny particle that holds an electrical charge. This charge can be either positive or negative. In most problems, these charges are considered to be fixed in position and are small enough that the distance between them can be treated as the defining factor of their interactions.

Point charges exert forces on each other. They can either attract or repel based on their charge. Like charges repel, while opposite charges attract. The force between any two point charges depends on their magnitudes and the distance separating them. In problems involving point charges, it is vital to pay attention to these attributes as they play crucial roles in calculating forces and potential energies.
Coulomb's Law
Coulomb's law is a fundamental principle in electrostatics. It describes how the force between two point charges is determined. The law can be expressed with the formula: \[ F = k \cdot \frac{|q_1 \cdot q_2|}{r^2} \]
In this formula:
  • \(F\) is the force between the charges.
  • \(k\) is Coulomb's constant, approximately \(8.99 \times 10^9 \text{ N m}^2/\text{C}^2\).
  • \(q_1\) and \(q_2\) represent the quantities of the charges.
  • \(r\) is the distance between the centers of the two charges.
Coulomb's law not only quantifies the magnitude of the force but also the direction. If the product \(q_1 \cdot q_2\) is positive, the force is repulsive. Conversely, if the product is negative, the force is attractive. This law is crucial in determining how point charges will behave in any electrostatic setup.
Equilateral Triangle
In the context of geometry, an equilateral triangle is a triangle where all three sides are equal in length. This symmetry is beneficial in problems involving electric charges because it simplifies calculations.

Each angle in an equilateral triangle measures \(60^\circ\). In our specific exercise, the charges are placed at each corner of the triangle, necessitating the calculation of potential energy between each pair of charges. On occasions where geometric symmetry, like that in an equilateral triangle, exists, calculations become more straightforward, and a comprehensive understanding of spatial arrangements aids in dissecting and solving these physics problems easily.
Electrostatics
Electrostatics is the branch of physics that studies electric charges at rest. It involves analyzing static electric forces, electric fields, and potential energy. It’s vital to include potential energy as it dictates how charges interact and whether they are in stable configurations or not.

When multiple charges exist in a system, like in our exercise, their arrangement determines the potential energy. This is derived from pairing every two charges and computing their potential energy. The overall potential energy of the system is the summation of all individual pair potentials. Understanding electrostatic interactions is key to mastering concepts related to charge distributions and their energy implications.

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 a certain region of space, the electric potential is \(V(x, y, z)=A x y-B x^{2}+C y,\) where \(A, B,\) and \(C\) are positive constants. (a) Calculate the \(x-y^{-},\) and \(z\) -components of the electric field. (b) At which points is the electric field equal to zero?

A very long insulating cylinder of charge of radius \(2.50 \mathrm{~cm}\) carries a uniform linear density of \(15.0 \mathrm{nC} / \mathrm{m}\). If you put one probe of a voltmeter at the surface, how far from the surface must the other probe be placed so that the voltmeter reads \(175 \mathrm{~V} ?\)

A particle with charge \(+4.20 \mathrm{nC}\) is in a uniform electric field \(\vec{E}\) directed to the left. The charge is released from rest and moves to the left; after it has moved \(6.00 \mathrm{~cm},\) its kinetic energy is \(+2.20 \times 10^{-6} \mathrm{~J}\). What are (a) the work done by the electric force, (b) the potential of the starting point with respect to the end point, and (c) the magnitude of \(\overrightarrow{\boldsymbol{E}}\) ?

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?

At a certain distance from a point charge, the potential and electric-field magnitude due to that charge are \(4.98 \mathrm{~V}\) and \(16.2 \mathrm{~V} / \mathrm{m}\) respectively. (Take \(V=0\) at infinity.) (a) What is the distance to the point charge? (b) What is the magnitude of the charge? (c) Is the electric field directed toward or away from the noint charge?

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.