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Four identical charges \((+2.0 \mu \mathrm{C}\) each \()\) are brought from infinity and fixed to a straight line. The charges are located \(0.40 \mathrm{m}\) apart. Determine the electric potential energy of this group.

Short Answer

Expert verified
The total electric potential energy is 0.39 J.

Step by step solution

01

Understand the Problem

We need to find the total electric potential energy for four identical point charges aligned in a straight line. Each charge is \(+2.0 \mu \text{C}\) and they are separated by \(0.40 \text{ m}\). The potential energy between each pair of charges must be calculated and summed up.
02

Calculate Potential Energy between a Pair of Charges

The formula to calculate the electric potential energy between two point charges \(q_1\) and \(q_2\) separated by a distance \(r\) is:\[ U = \frac{k \cdot q_1 \cdot q_2}{r} \]where \(k = 8.99 \times 10^9 \text{ N m}^2/\text{C}^2\) is Coulomb's constant. For our problem, each charge \(q = +2.0 \times 10^{-6} \text{ C}\) and the distance \(r = 0.40 \text{ m}\).
03

Compute Individual Pair Energies

Using the formula, compute the potential energy for each pair of charges:- Energy between Charge 1 and Charge 2: \[ U_{12} = \frac{(8.99 \times 10^9) \cdot (2.0 \times 10^{-6})^2}{0.40} \approx 0.09 \text{ J} \]- Using the same process, this calculation applies to each adjacent pair. There are three adjacent pairs: Charge 1-2, Charge 2-3, and Charge 3-4, each with \(0.09 \text{ J}\).
04

Calculate Energies for Non-Adjacent Pairs

Next, calculate the potential energies for non-adjacent pairs:- Energy between Charge 1 and Charge 3 (distance = 0.80 m): \[ U_{13} = \frac{(8.99 \times 10^9) \cdot (2.0 \times 10^{-6})^2}{0.80} \approx 0.045 \text{ J} \]- Energy between Charge 2 and Charge 4 (distance = 0.80 m): \[ U_{24} = 0.045 \text{ J} \]- Energy between Charge 1 and Charge 4 (distance = 1.20 m): \[ U_{14} = \frac{(8.99 \times 10^9) \cdot (2.0 \times 10^{-6})^2}{1.20} \approx 0.03 \text{ J} \]
05

Sum All Potential Energies

Add all the calculated potential energies to find the total electric potential energy:\[ U_{total} = U_{12} + U_{23} + U_{34} + U_{13} + U_{24} + U_{14} \]\[ U_{total} = 0.09 + 0.09 + 0.09 + 0.045 + 0.045 + 0.03 = 0.39 \text{ J} \]
06

Conclusion

The total electric potential energy of the system of four charges aligned with specific distances is obtained by summing the energies of all possible charge pairs. This value represents the work required to assemble this configuration from infinity.

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Key Concepts

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

Coulomb's Law
Coulomb's Law is a formula used to describe the electric force between two charged objects. It helps us understand how much these objects will push or pull on each other. This law is particularly useful when calculating the force—and by extension, the electric potential energy—between point charges. In essence, Coulomb's Law states that the force between two point charges varies directly with the product of their magnitudes and inversely with the square of the distance between them. However, in our exercise, we are more focused on electric potential energy rather than force.

For electric potential energy, Coulomb's Law allows us to calculate the potential energy between two charges as \[ U = \frac{k \cdot q_1 \cdot q_2}{r} \]where \(q_1\) and \(q_2\) are the magnitudes of the charges, \(r\) is the distance that separates them, and \(k\) is a constant known as Coulomb's Constant.

The importance of Coulomb's Law in calculating electric potential energy lies in its ability to model individual interactions within a system of multiple charges. By doing so, we can sum these interactions to obtain the total potential energy of the system. Understanding the contributions of each pair of charges helps us get a complete picture of the energy stored in the arrangement.
Point Charges
A point charge refers to an idealized charge with no physical size; it's essentially treated as a mathematical point. This abstraction helps us simplify problems involving electric fields and forces, as it allows us to focus solely on the charge rather than its size or shape. In realistic scenarios, like in our exercise, point charges are considered to be perfectly spherical. As a result, the distance in the formula is measured from the centers of the charges.

This abstraction is valuable for calculations involving multiple charges, such as in the arrangement of four identical charges in the exercise. When calculating electric potential energy, each pair of point charges contributes based on their separation distance and magnitude. The simplicity of the point charge model makes it easier to compute and sum potential energies between multiple pairs, leading to a straightforward calculation of the system's total electric potential energy.
  • Each charge represents an interaction center.
  • Distances between centers define separation in equations.
The concept of point charges is widely applicable in physics to simplify complex situations, reducing them to calculations based solely on proximities and magnitudes.
Coulomb's Constant
Coulomb's Constant, often symbolized as \(k\), is a foundational constant in electrostatics. Its value is approximately \(8.99 \times 10^9 \text{ N m}^2/\text{C}^2\). This constant plays a crucial role in Coulomb's Law, influencing both the electric force between charges and the electric potential energy. It essentially quantifies the strength of the interaction per unit charge product and unit distance squared.

In the exercise, Coulomb’s Constant ensures that the calculations of electric potential energy are scaled correctly to reflect physical realities. Every time we calculate the potential energy between two charges, this constant translates the minute coulombs and sub-meter distances into joules of potential energy.
  • Links electrostatic force and charge interaction strength.
  • Applies universally for any point charges in a vacuum.
Understanding \(k\) helps contextualize how powerful electric forces and potential energies can be, highlighting why charged particles significantly affect each other even at relatively large distances as shown in our exercise's aligned charge setup.

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Most popular questions from this chapter

An electron and a proton are initially very far apart (effectively an infinite distance apart). They are then brought together to form a hydrogen atom, in which the electron orbits the proton at an average distance of \(5.29 \times\) \(10^{-11} \mathrm{m} .\) What is \(\mathrm{EPE}_{\text {final }}-\mathrm{EPE}_{\text {initial }},\) which is the change in the electric potential energy?

Charges of \(-q\) and \(+2 q\) are fixed in place, with a distance of \(2.00 \mathrm{m}\) between them. A dashed line is drawn through the negative charge, perpendicular to the line between the charges. On the dashed line, at a distance \(L\) from the negative charge, there is at least one spot where the total potential is zero. Find \(L\).

Identical \(+1.8 \mu \mathrm{C}\) charges are fixed to adjacent corners of a square. What charge (magnitude and algebraic sign) should be fixed to one of the empty corners, so that the total electric potential at the remaining empty corner is \(0 \mathrm{V} ?\)

One particle has a mass of \(3.00 \times 10^{-3} \mathrm{kg}\) and a charge of \(+8.00 \mu \mathrm{C}\). A second particle has a mass of \(6.00 \times 10^{-3} \mathrm{kg}\) and the same charge. The two particles are initially held in place and then released. The particles fly apart, and when the separation between them is \(0.100 \mathrm{m},\) the speed of the \(3.00 \times 10^{-3} \mathrm{kg}\) particle is \(125 \mathrm{m} / \mathrm{s} .\) Find the initial separation between the particles.

Particle 1 has a mass of \(m_{1}=3.6 \times 10^{-6} \mathrm{kg},\) while particle 2 has a mass of \(m_{2}=6.2 \times 10^{-6} \mathrm{kg} .\) Each has the same electric charge. These particles are initially held at rest, and the two- particle system has an initial electric potential energy of 0.150 J. Suddenly, the particles are released and fly apart because of the repulsive electric force that acts on each one (see the figure). The effects of the gravitational force are negligible, and no other forces act on the particles. Concepts: (i) What types of energy does the twoparticle system have initially? (ii) What types of energy does the two-particle system have at the instant illustrated in part \(b\) of the drawing? (iii) Does the principle of conservation of energy apply to this problem? Explain. (iv) Does the conservation of linear momentum apply to the two particles as they fly apart? Explain. Calculations: At one instant following the release, the speed of particle 1 is measured to be \(v_{1}=170 \mathrm{m} / \mathrm{s}\). What is the electric potential energy at this instant? (a) Two particles have different masses, but the same electrical charge \(q\) They are initially at rest. (b) At the instant following the release of the particles, they are flying apart due to the mutual force of electric repulsion.

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