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Each of the following pairs of charges are separated by a distance \(d\). Which pair has the highest potential energy? a) \(+5 \mathrm{C}\) and \(+3 \mathrm{C}\) b) \(+5 \mathrm{C}\) and \(-3 \mathrm{C}\) c) \(-5 \mathrm{C}\) and \(+3 \mathrm{C}\) d) All pairs have the same potential energy.

Short Answer

Expert verified
Answer: Pair (a) has the highest potential energy.

Step by step solution

01

Analyze Pair (a)

Pair (a) consists of two positive charges: \(+5\mathrm{C}\) and \(+3\mathrm{C}\). Using the formula for potential energy, we substitute the values: \(U_a = k \dfrac{(5)(3)}{d} = k\dfrac{15}{d}\)
02

Analyze Pair (b)

Pair (b) consists of a positive charge and a negative charge: \(+5\mathrm{C}\) and \(-3\mathrm{C}\). Substitute the values into the potential energy formula: \(U_b = k \dfrac{(5)(-3)}{d} = k\dfrac{-15}{d}\)
03

Analyze Pair (c)

Pair (c) is composed of a negative charge and a positive charge: \(-5\mathrm{C}\) and \(+3\mathrm{C}\). Plugging the values into the potential energy formula: \(U_c = k \dfrac{(-5)(3)}{d} = k\dfrac{-15}{d}\)
04

Compare the Potential Energies

Now that we have calculated the potential energy for each pair, we need to compare them to determine which one is the highest: \(U_a = k\dfrac{15}{d}\) \(U_b = k\dfrac{-15}{d}\) \(U_c = k\dfrac{-15}{d}\) The potential energy of pair (a) is \(k\dfrac{15}{d}\), while the potential energy of pairs (b) and (c) is negative. So, we can conclude that pair (a) has the highest potential energy. Therefore, the correct answer is pair (a).

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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 fundamental principle in electrostatics that describes the force between two charged objects. Formulated by Charles-Augustin de Coulomb in the 18th century, it states that the magnitude of the electrostatic force between two point charges is directly proportional to the product of the charges and inversely proportional to the square of the distance separating them.

This relationship can be expressed by the formula:
\[\begin{equation}F = k \frac{|q_1 q_2|}{r^2},\text{where:}\begin{itemize} item \(F\) is the force between charges, item \(k\) is Coulomb's constant ( 8.987 \times 10^9 N \text{m}^2\text{C}^{-2}), item \(q_1\) and \(q_2\) are the amounts of the charges, item \(r\) is the separation distance between the center of the charges,\text{and} item The force is attractive if the charges are of opposite signs and repulsive if they are the same.\text{end}{itemize}\text{end}{equation}
Understanding Coulomb's Law is key to solving a wide array of physics problems, including those that involve electrostatic potential energy. The formula for electrostatic potential energy (or electric potential energy) between two point charges is derived from Coulomb's Law and is given by:
\[\begin{equation}U = k \frac{q_1 q_2}{r},\text{end}{equation}
where \(U\) represents the potential energy. This is the formula we apply when determining the potential energy of charged object pairs, as seen in the textbook exercise.
Electric Charge
Electric charge is a fundamental property of matter that causes it to experience a force when placed in an electromagnetic field. There are two types of electric charge, commonly known as positive and negative charges. Charges that are the same repel each other, while opposite charges attract. This interaction is the basis of the electric force that acts between charged objects.

The SI unit of charge is the Coulomb (C), indicating the quantity of electricity conveyed by a current of one ampere flowing for one second. One Coulomb is roughly equivalent to 6.242 x 10^18 times the charge of a single electron or proton.

Considering the textbook exercise, it’s important to note that the magnitude of charges, whether positive or negative, influences the resulting potential energy in a situation. The electric charge impacts not only the magnitude but also the sign of electrostatic potential energy between charges: a positive result implies repulsion, and a negative one indicates attraction.
Physics Problem Solving
Physics problem solving is a critical skill that requires understanding concepts, knowing how to apply mathematical formulas, and logical reasoning. When solving physics problems, especially relating to electrostatics, it's essential to follow a systematic approach:

1. Identify the physical principles involved. In the case of our textbook problem, we use the electrostatic potential energy concept derived from Coulomb's Law.
2. Translate the physical situation into a mathematical model. Here, we express the potential energies of the charge pairs with the formula given by Coulomb’s Law.
3. Execute the necessary mathematical operations. By substituting values into the potential energy formula, we obtain the potential energy of each charge pair.
4. Compare and analyze the results. In the final step, we compared the values, ensuring the proper interpretation of signs and magnitude for electric charges.

Correct interpretation of positive and negative signs is crucial, as they can indicate the nature of the force between charges - attractive or repulsive. Improving upon the existing textbook solution approach can include discussing common misconceptions, such as overlooking that charges can be negative or misunderstanding the significance of potential energy's sign.

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Most popular questions from this chapter

A solid conducting sphere of radius \(R_{1}=1.435 \mathrm{~m}\) has a charge of Q evenly distributed over its surface. A second solid conducting sphere of radius \(R_{2}=0.6177 \mathrm{~m}\) is initially uncharged and at a distance of \(10.00 \mathrm{~m}\) from the first sphere. The two spheres are momentarily connected with a wire, which is then removed. The resulting charge on the second sphere is \(0.9356 \mu \mathrm{C}\). What was the original charge, \(Q\), on the first sphere?

A positive charge is released and moves along an electric field line. This charge moves to a position of a) lower potential and lower potential energy. b) lower potential and higher potential energy. c) higher potential and lower potential energy. d) higher potential and higher potential energy.

Which of the following angles between an electric dipole moment and an applied electric field will result in the most stable state? a) \(0 \mathrm{rad}\) b) \(\pi / 2 \mathrm{rad}\) c) \(\pi \mathrm{rad}\) d) The electric dipole moment is not stable under any condition in an applied electric field.

The electric field, \(\vec{E}(\vec{r}),\) and the electric potential, \(V(\vec{r}),\) are calculated from the charge distribution, \(\rho(\vec{r}),\) by integrating Coulomb's Law and then the electric field. In the other direction, the field and the charge distribution are determined from the potential by suitably differentiating. Suppose the electric potential in a large region of space is given by \(V(r)=V_{0} \exp \left(-r^{2} / a^{2}\right),\) where \(V_{0}\) and \(a\) are constants and \(r=\sqrt{x^{2}+y^{2}+z^{2}}\) is the distance from the origin. a) Find the electric field \(\vec{E}(\vec{r})\) in this region. b) Determine the charge density, \(\rho(\vec{r}),\) in this region, which gives rise to the potential and field. c) Find the total charge in this region. d) Roughly sketch the charge distribution that could give rise to such an electric field.

A point charge of \(+2.00 \mu \mathrm{C}\) is located at \((2.50 \mathrm{~m}, 3.20 \mathrm{~m}) .\) A second point charge of \(-3.10 \mu C\) is located at \((-2.10 \mathrm{~m}, 1.00 \mathrm{~m})\) a) What is the electric potential at the origin? b) Along a line passing through both point charges, at what point(s) is (are) the electric potential(s) equal to zero?

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