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Chapter 18: Problem 9

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 \( 0.0777 \, \mathrm{J} \).

Step by step solution

01

Understand the Formula for Potential Energy of a System of Point Charges

The formula to calculate the potential energy \( U \) of a system of point charges is given by \( U = k \cdot \frac{q_1 q_2}{r} \), where \( k \) is Coulomb's constant \( 8.99 \times 10^9 \, \mathrm{Nm^2/C^2} \), \( q_1 \) and \( q_2 \) are the charges, and \( r \) is the distance between the charges.
02

Determine the Pairs of Charges

In an equilateral triangle with three charges, there are three unique pairs of charges: AB, BC, and CA. We need to compute the potential energy for each pair.
03

Calculate Potential Energy for One Pair of Charges

For a pair of charges with \( q = 1.20 \, \mu \mathrm{C} = 1.20 \times 10^{-6} \, \mathrm{C} \) and a distance \( r = 0.500 \, \mathrm{m} \), the potential energy \( U_{AB} \) is calculated as:\[ U_{AB} = k \cdot \frac{q_1 q_2}{r} = (8.99 \times 10^9 \, \mathrm{Nm^2/C^2}) \cdot \frac{(1.20 \times 10^{-6} \, \mathrm{C})^2}{0.500 \, \mathrm{m}} \].
04

Compute Potential Energy for All Pairs

Since the triangle is equilateral, the potential energy \( U \) for each pair (AB, BC, CA) is the same. Therefore, compute:\[ U_{AB} = U_{BC} = U_{CA} \]. Then, multiply by 3 (the number of pairs) to find the total potential energy of the system:\[ U_{total} = 3 \times U_{AB} \].
05

Substitute Values and Solve

Substitute the given values into the formula:\[ U_{AB} = (8.99 \times 10^9) \cdot \frac{(1.20 \times 10^{-6})^2}{0.500} = 2.59 \times 10^{-2} \, \mathrm{J} \].Now, calculate \( U_{total} \):\[ U_{total} = 3 \times 2.59 \times 10^{-2} = 7.77 \times 10^{-2} \, \mathrm{J} \].

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Coulomb's Law: Understanding Electrostatic Forces
Coulomb's Law is fundamental for understanding electrostatic forces between point charges. The law describes the interaction between two electrically charged particles. It states that the force between them is directly proportional to the product of their charges and inversely proportional to the square of the distance between their centers. The formula is written as:\[ F = k \cdot \frac{|q_1 q_2|}{r^2} \]where:
  • F is the electrostatic force between the charges.
  • k is the Coulomb's constant, approximately \(8.99 \times 10^9 \, \mathrm{Nm^2/C^2} \).
  • q1 and q2 are the magnitudes of the charges.
  • r is the distance between the charges.

It's essential to remember that this law only applies to point charges or spherical charge distributions when considering electrostatics in a vacuum. It explains why particles with like charges repel each other and opposite charges attract.
System of Point Charges: Combining Their Energies
A system of point charges involves multiple charged particles where the potential energy is calculated between each pair of charges. In our exercise, three point charges form such a system, each interacting with the others.The potential energy between two charges in this system is given by the formula:\[ U = k \cdot \frac{q_1 q_2}{r} \]where U is the potential energy, and the other variables are defined as in Coulomb's Law. The total potential energy of the system is the sum of energies from all unique pairs of charges.

For a system with three charges, like in the equilateral triangle:
  • There are three pairs: AB, BC, and CA.
  • The potential energy for each pair is calculated using the formula above.
  • The total energy is the sum of these individual energies.

This computation assumes no external electric fields are affecting the charges, allowing us to focus only on the interactions within the system.
Equilateral Triangle Charges: Symmetry and Calculations
When point charges are arranged in an equilateral triangle, symmetry simplifies calculations. In this geometry, not only do all charges have the same magnitude, but they also lie at equal distances from each other. This specific arrangement allows:
  • Equidistant spacing, making the calculations for potential energy identical for each pair of charges.
  • Uniform distribution of charge interactions, which is handy in simplifying the computation of total potential energy.

The side length of 0.500 meters means each pair of charges is separated by the same distance. Given the uniform distances, the potential energy calculated for one pair can be multiplied by the number of pairs (three in the case of our triangle) to quickly find total energy. This concept is crucial in advanced physics where symmetrical configurations often allow for elegant solutions and deeper insights.

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Most popular questions from this chapter

Electrical sensitivity of sharks. Certain sharks can detect an electric field as weak as 1.0\(\mu \mathrm{V} / \mathrm{m} .\) To grasp how weak this field is, if you wanted to produce it between two parallel metal plates by connecting an ordinary 1.5 \(\mathrm{V}\) A battery across theseplates, how far apart would the plates have to be?

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 \(E (\mathrm{c})\) A negative point charge \(q=-0.200 \mu \mathrm{C}\) is moved from \(b\) to \(a\) . Calculate the work done on the point charge by the electric field.

How much charge does a 12 \(\mathrm{V}\) battery have to supply to fully charge a 2.5\(\mu \mathrm{F}\) capacitor and a 5.0\(\mu \mathrm{F}\) capacitor when they're (a) in parallel, (b) in series? (c) How much energy does the battery have to supply in each case?

When two point charges are a distance \(R\) apart, their potential energy is \(-2.0 \mathrm{J} .\) How far far (in terms of \(R )\) should they be from each other so that their potential energy is \(-6.0 \mathrm{J} ?\)

A point charge \(Q=+4.60 \mu \mathrm{C}\) is held fixed at the origin. A second point charge \(q=+1.20 \mu \mathrm{C}\) with mass of \(2.80 \times\) \(10^{-4} \mathrm{kg}\) is placed on the \(x\) axis, 0.250 \(\mathrm{m}\) from the origin. (a) What is the electric potential energy \(U\) of the pair of charges? (Take \(U\) to be zero when the charges have infinite separation.) (b) The second point charge is released from rest. What is its speed when its distance from the origin is (i) \(0.500 \mathrm{m} ;\) (ii) 5.00 \(\mathrm{m}\) ; (iii) 50.0 \(\mathrm{m} ?\)
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