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Chapter 19: Problem 20

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 of this group is 0.2685 J.

Step by step solution

01

Identify the Problem

We want to find the total electric potential energy of four identical charges arranged linearly. Each charge is +2.0 碌C and they're 0.40 meters apart.
02

Understanding Electric Potential Energy

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

Calculate Pairwise Potential Energy

Since we have four charges, we need to calculate the potential energy for each pair of charges. The potential energy for each pair with distance \(r\) can be calculated and summed for all pairs.
04

List All Charge Pairs and Distances

Arrange the charges as A, B, C, D. Then, calculate the potential energy between pairs: \(AB\) (same for \(BC\) and \(CD\)), distance = 0.40 m; \(AC\), distance = 2 \(\times\) 0.40 m; \(AD\) and \(BD\), distance = 3 \(\times\) 0.40 m; \(BC\), \(CD\), distance = 0.40 m.
05

Calculate Energies for 0.40 m Pairs

The potential energy between charges \((AB, BC, CD)\) is \(U_{AB} = U_{BC} = U_{CD} = \frac{k \times (+2.0 \times 10^{-6})^2}{0.40} = 0.0896 \, ext{J}\).
06

Calculate Energies for 0.80 m Pairs

The potential energy between charges \((AC, BD)\) is \(U_{AC} = U_{BD} = \frac{k \times (+2.0 \times 10^{-6})^2}{0.80} = 0.0448 \, ext{J}\).
07

Calculate Energy for 1.20 m Pair

The potential energy between the charges \((AD)\) is \(U_{AD} = \frac{k \times (+2.0 \times 10^{-6})^2}{1.20} = 0.0299 \, ext{J}\).
08

Sum All Potential Energies

Add the potential energies from all relevant pairs to find the total potential energy: \(3 \times 0.0896 + 2 \times 0.0448 + 1 \times 0.0299 = 0.2685 \, ext{J}\).

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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 crucial for understanding electric forces between charged objects. It describes how two point charges interact with each other. The law states that the magnitude of the force \(F\) between two point charges, \(q_1\) and \(q_2\), is directly proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance \(r\) between them. The formula is:\[F = k \frac{|q_1 q_2|}{r^2}\]where \(k\) is Coulomb's constant, approximately \(8.99 \times 10^9 \, \text{N m}^2/\text{C}^2\). This law helps understand both the direction and magnitude of the force. If both charges are of the same sign, they repel each other, while opposite charges attract. In the problem of four charges on a line, Coulomb's Law lets us calculate each pairwise interaction's electric potential energy.
Point Charges
Point charges are idealized charges considered to have no dimensions, meaning their size is negligible compared to the distances between them. This simplification allows us to apply Coulomb's Law straightforwardly. Real charges have size and structure, but assuming them to be point charges makes calculations feasibleand helps learn core concepts.
  • Point charges can be positive or negative.
  • They interact via electric forces in empty space.
  • Potential energy between two point charges is determined by their magnitudes and separation distance.
For the given exercise with four identical charges of \(+2.0 \, \mu \text{C}\), considering them as point charges simplifies the calculation of electric potential energy by focusing solely on charge values and distances.
Linear Charge Configuration
A linear charge configuration is when multiple charges are arranged in a straight line. It is a simplified model in electrostatics to study interactions between multiple charges. Calculating the total electric potential energy for such a configuration involves accounting for every possible pair of interactions. In this scenario, understanding the following will make these calculations easier:
  • Every adjacent pair of charges has a different distance than non-adjacent pairs.
  • Non-adjacent pairs contribute significantly less to the sum of potential energy because of larger distances.
  • The symmetry of a linear arrangement can sometimes simplify calculations.
To solve the given problem, each pair of charges along the line contributes to the total electric potential energy. Distances like \(0.40 \text{ m}, 0.80 \text{ m},\) and \(1.20 \text{ m}\) were utilized to compute pairwise potential energies, which were then summed to find the system's total electric potential energy.

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Most popular questions from this chapter

A capacitor stores \(5.3 \times 10^{-5} \mathrm{C}\) of charge when connected to a 6.0 -V battery. How much charge does the capacitor store when connected to a \(9.0-\mathrm{V}\) battery?

Concept Questions 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 \(v_{\mathrm{e}},\) while the proton acquires a speed \(v_{\mathrm{p}}\). (a) As each particle accelerates from rest, it gains kinetic energy. Does it gain or lose electric potential energy? (b) Does the electron gain more, less, or the same amount of kinetic energy as the proton does? (c) Is \(v_{\mathrm{e}}\) greater than, less than, or equal to \(v_{\mathrm{p}}\) ? Justify your answers. Find the ratio \(v_{\mathrm{e}} / v_{\mathrm{p}}\). Verify that your answer is consistent with your answers to the Concept Questions.

A positive point charge is surrounded by an equipotential surface \(A\), which has a radius of \(r_{A}\). A positive test charge moves from surface \(A\) to another equipotential surface \(B\), which has a radius of \(r_{B}\). In the process, the electric force does negative work. (a) Does the electric force acting on the test charge have the same or opposite direction as the displacement of the test charge? (b) Is \(r_{B}\) greater than or less than \(r_{A}\) ? Explain your answers. Problem The positive point charge is \(q=+7.2 \times 10^{-8} \mathrm{C}\), and the test charge is \(q_{0}=+4.5 \times 10^{-11} \mathrm{C}\). The work done as the test charge moves from surface \(A\) to surface \(B\) is \(W_{A B}=-8.1 \times 10^{-9} \mathrm{~J}\). The radius of surface \(A\) is \(r_{A}=1.8 \mathrm{~m}\). Find \(r_{B}\). Check to see that your answer is consistent with your answers to the Concept Questions.

Multiple-Concept Example 4 deals with the concepts that are important in this problem. As illustrated in Figure \(19-6 b\), a negatively charged particle is released from rest at point \(B\) and accelerates until it reaches point \(A\). The mass and charge of the particle are \(4.0 \times 10^{-6} \mathrm{~kg}\) and \(-2.0 \times 10^{-5} \mathrm{C},\) respectively. Only the gravitational force and the electrostatic force act on the particle, which moves on a horizontal straight line without rotating. The electric potential at \(A\) is \(36 \mathrm{~V}\) greater than that at \(B\); in other words, \(V_{A}-V_{B}=36 \mathrm{~V}\). What is the translational speed of the particle at point \(A ?\)

Refer to Multiple-Concept Example 3 to review the concepts that are needed here. A cordless electric shaver uses energy at a rate of \(4.0 \mathrm{~W}\) from a rechargeable \(1.5-\mathrm{V}\) battery. Each of the charged particles that the battery delivers to the shaver carries a charge that has a magnitude of \(1.6 \times 10^{-19} \mathrm{C}\). A fully charged battery allows the shaver to be used for its maximum operation time, during which \(3.0 \times 10^{22}\) of the charged particles pass between the terminals of the battery as the shaver operates. What is the shaver's maximum operation time?
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