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Three equal 1.20-\(\mu$$C\) point charges are placed at the corners of an equilateral triangle with sides 0.400 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 0.0971 J.

Step by step solution

01

Understanding the Problem

We need to find the potential energy of a system of three equal point charges arranged in an equilateral triangle. Each charge is \(1.20\,\mu C\), and the side length of the triangle is \(0.400\,m\). The potential energy of the system should be calculated relative to when the charges are infinitely far apart.
02

Formula for Potential Energy Between Two Charges

The potential energy between two point charges \(q_1\) and \(q_2\) separated by a distance \(r\) is given by \(U = \frac{k \cdot q_1 \cdot q_2}{r}\), where \(k = 8.99 \times 10^9 \, N\cdot m^2/C^2\) is Coulomb's constant.
03

Calculate Pairwise Potential Energy

Since the charges are equal, each pair of charges contributes the same amount to the total potential energy. Calculate the potential energy for one pair of charges, \(U_{pair} = \frac{(8.99 \times 10^9) \cdot (1.20 \times 10^{-6})^2}{0.400}\).
04

Compute the Contribution for All Pairs

There are three pairs of charges in an equilateral triangle. Multiply the pairwise potential energy from the last step by 3 to find the total potential energy. \(U_{total} = 3 imes U_{pair}\).
05

Substitute Values and Solve

Substitute the given values into the formula: \(U_{pair} = \frac{8.99 \times 10^9 \cdot 1.44 \times 10^{-12}}{0.400} = 3.2376 \times 10^{-2} \). Then, calculate the total potential energy: \(U_{total} = 3 \times 3.2376 \times 10^{-2} = 9.7128 \times 10^{-2} \, J\).

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Key Concepts

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

Potential Energy
Potential energy in electrostatics is the energy stored in a system of charged particles due to their positions and interactions. When dealing with point charges, the potential energy of the system arises from the electrostatic forces between these charges.
To calculate the potential energy between two point charges, we use the formula:
  • \( U = \frac{k \cdot q_1 \cdot q_2}{r} \)
Here, \(U\) is the potential energy, \(k\) is Coulomb's constant, \(q_1\) and \(q_2\) are the magnitudes of the charges, and \(r\) is the distance between them. In our exercise, the potential energy is calculated for an arrangement of charges in an equilateral triangle. We find the energy required to bring each charge into position from infinitely far away, where their potential energy is zero.
By summing the potential energy contributions of each pair of charges, we obtain the total potential energy of the system. Also, since the charges are identically arranged, the calculation simplifies to multiplying the energy of one charge pair by the number of pairs.
Coulomb's Constant
Coulomb's constant, denoted as \(k\), is a fundamental constant in electrostatics. It plays a crucial role in calculating the electric force and potential energy between point charges. Its value is approximately \(8.99 \times 10^9 \, ext{N} \cdot ext{m}^2/ ext{C}^2\).
Coulomb's constant arises from experiments that measure the force exerted between charged particles. It is used in Coulomb's Law, which states that the electric force between two point charges is directly proportional to the product of their charges and inversely proportional to the square of the distance between them. Similarly, in potential energy calculations, \(k\) shows the intensity of the interaction between charges.
The significance of \(k\) helps us understand how strong or weak the interaction between two charges is at a given distance. A large \(k\) indicates a strong interaction, which is crucial in systems like the one in the exercise where charges are closely packed in a triangle.
Point Charges
Point charges are hypothetical charges located at a single point in space. This simplification allows us to analyze charge interactions without considering the size or shape of the objects holding the charges. It aids in understanding the fundamental behaviors of charges.
In the exercise, the charges are all equal and identical, each having a magnitude of \(1.20 \, \mu C\). This uniformity simplifies the calculation of potential energy because the changes between any pair of charges are the same.
Point charges are an idealization; in reality, objects have size and structure. However, by considering them as point charges, we're able to apply mathematical models readily and focus solely on the effects of charge and distance. This is why in problems like the one provided, point charges are a common representation to study the principles of electrostatics.
Equilateral Triangle
An equilateral triangle is a triangle where all three sides are of equal length. In our problem, this means the distance between any two charges is the same. Each side is given as \(0.400 \, m\). This property simplifies calculations because it ensures uniformity in the electric and potential energies being calculated.
Due to the symmetry of an equilateral triangle, every charge affects every other charge equally. Thus, each pair of charges in the triangle contributes evenly to the total potential energy.
This geometric arrangement also makes it easier to apply formulas consistently since the same distance \(r\) can be used in each calculation. Therefore, understanding that the triangle is equilateral is key in recognizing the repetitive and consistent calculations needed for potential energy in the given system of charges.

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Most popular questions from this chapter

The electric field at the surface of a charged, solid, copper sphere with radius 0.200 m is 3800 N\(/\)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?

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A small particle has charge \(-5.00\) \(\mu\)C and mass \(2.00 \times 10^{-4}\) kg. It moves from point \(A\), where the electric potential is \(V_A = +\)200 V, to point \(B\), where the electric potential is \(V_B = +\)800 V. The electric force is the only force acting on the particle. The particle has speed 5.00 m\(/\)s at point \(A\). What is its speed at point \(B\)? Is it moving faster or slower at \(B\) than at \(A\)? Explain.

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