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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 \(0.953 \mathrm{J}\).

Step by step solution

01

Understanding the Problem

We need to find the electric potential energy of a system of four identical point charges, each with a charge of \(+2.0 \mu \mathrm{C}\). These charges are aligned in a straight line and are separated by a distance of \(0.40\, \mathrm{m}\) from each other.
02

Formula for Electric Potential Energy

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

Calculate Pairwise Energies

Since there are four charges, we need to calculate the potential energy for each pair of charges:1. \(U_{12} = \frac{k \cdot q \cdot q}{0.40} = \frac{8.99 \times 10^9 \cdot (2.0 \times 10^{-6})^2}{0.40}\)2. \(U_{13} = \frac{k \cdot q \cdot q}{0.80} = \frac{8.99 \times 10^9 \cdot (2.0 \times 10^{-6})^2}{0.80}\)3. \(U_{14} = \frac{k \cdot q \cdot q}{1.20} = \frac{8.99 \times 10^9 \cdot (2.0 \times 10^{-6})^2}{1.20}\)4. Similarly, calculate other combinations for total pairs.
04

Calculate Total Potential Energy

After calculating the potential energy for each unique pair of charges (\(U_{12}, U_{13}, U_{14}, U_{23}, U_{24}, U_{34}\)), sum all these energies to find the total potential energy of the system.
05

Evaluate and Sum the Energies

Using the distances, the energies are calculated as follows:\(U_{12} = 0.22 J, U_{13} = 0.11 J, U_{14} = 0.073 J\)\(U_{23} = 0.22 J, U_{24} = 0.11 J, U_{34} = 0.22 J\)Add all these energies to find the total:\[ U_{total} = 0.22 + 0.11 + 0.073 + 0.22 + 0.11 + 0.22 = 0.953 J \]
06

Conclusion

The total electric potential energy of the four-charge system is \(0.953 \mathrm{J}\).

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Key Concepts

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

Point Charges
In electrostatics, point charges are theoretical charges with no physical size or volume. This concept simplifies the calculations because it allows us to treat charges as discrete entities rather than spread-out distributions. Point charges are often used in physics to model atoms or ions when analyzing electric fields or potential energies.
By treating charges as points, it becomes easier to apply mathematical formulas to solve problems involving electric potential energy and force interactions. Particularly in the case of multiple charges aligned on a straight line, it enables us to calculate the energy interactions systematically.
In the exercise, the point charges each have a magnitude of \(+2.0 \mu C\) and are separated by a distance of \(0.40\, \mathrm{m}\) between adjacent charges, which is crucial for calculating the pairwise electric potential energy between them.
Electrostatic Constant
The electrostatic constant, denoted by \(k\), plays a vital role in calculating the electric potential energy between point charges. The constant is a value derived from Coulomb's law, which quantifies the force of attraction or repulsion between two electric charges. In SI units, the electrostatic constant \(k\) is approximately \(8.99 \times 10^9 \, N\cdot m^2/C^2\).
This constant helps us understand how charges interact over a distance. When using the formula for electric potential energy \( U = \frac{k \cdot q_1 \cdot q_2}{r} \), \(k\) determines how strong the interaction is based on their separation and magnitudes. Higher values of \(k\) indicate stronger interactions at large distances.
  • Coulomb's law attributes this constant to enhance our understanding of electric forces.
  • In practice, this allows us to compute accurate potential energy values in multi-charge systems.
Pairwise Potential Energy Calculations
Pairwise potential energy calculations are essential for understanding how multiple point charges interact within a system. In such calculations, you consider each pair of charges separately and determine the potential energy between them using the formula \( U = \frac{k \cdot q_1 \cdot q_2}{r} \). This step-by-step approach helps break down complex scenarios into manageable calculations.
For the given exercise, with multiple point charges along a line, you identify each unique pair, compute the potential energy, and sum them up to find the total potential energy of the system. This involves:
  • Identifying each pair like \(U_{12}\), \(U_{13}\), etc.
  • Calculating the energy for each pair using their respective distances like \(0.40\, \mathrm{m}\), \(0.80\, \mathrm{m}\), etc.
  • Summing up all pairwise energies to determine the total energy.
This methodical approach simplifies the aggregation of potential energy in systems with more than two charges, ensuring clarity and accuracy in computations.

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Most popular questions from this chapter

An electron and a proton, starting from rest, are accelerated through an electric potential difference of the same magnitude. In the process, the electron acquires a speed \(v_{e},\) while the proton acquires a speed \(v_{\mathrm{p}} .\) Find the ratio \(v_{\mathrm{e}} / v_{\mathrm{p}}\).

An electron is released from rest at the negative plate of a parallel plate capacitor and accelerates to the positive plate (see the drawing). The plates are separated by a distance of \(1.2 \mathrm{cm},\) and the electric field within the capacitor has a magnitude of 2.1 \(\times 10^{6} \mathrm{V} / \mathrm{m}\). What is the kinetic energy of the electron just as it reaches the positive plate?

A positive charge \(q_{1}\) is located \(3.00 \mathrm{m}\) to the left of a negative charge \(q_{2} .\) The charges have different magnitudes. On the line through the charges, the net electric field is zero at a spot \(1.00 \mathrm{m}\) to the right of the negative charge. On this line there are also two spots where the total electric potential is zero. Locate these two spots relative to the negative charge.

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 (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. 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 two-particle 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?

Equipotential surface \(A\) has a potential of \(5650 \mathrm{V},\) while equipotential surface \(B\) has a potential of 7850 V. A particle has a mass of \(5.00 \times 10^{-2} \mathrm{kg}\) and a charge of \(+4.00 \times 10^{-5} \mathrm{C} .\) The particle has a speed of \(2.00 \mathrm{m} / \mathrm{s}\) on surface \(A .\) A nonconservative outside force is applied to the particle, and it moves to surface \(B\), arriving there with a speed of \(3.00 \mathrm{m} / \mathrm{s}\). How much work is done by the outside force in moving the particle from \(A\) to \(B ?\)

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