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Chapter 24: Problem 68

Here are the charges and coordinates of two point charges located in an \(x y\) plane: \(q_{1}=+3.00 \times 10^{-6} \mathrm{C}, x=+3.50 \mathrm{~cm}\), \(y=+0.500 \mathrm{~cm}\) and \(q_{2}=-4.00 \times 10^{-6} \mathrm{C}, x=-2.00 \mathrm{~cm}, y=+1.50\) \(\mathrm{cm}\). How much work must be done to locate these charges at their given positions, starting from infinite separation?

Short Answer

Expert verified
About 19.30 Joules of work is needed to position the charges.

Step by step solution

01

Understand the Concept

Work required to bring charges from infinity to a separation distance is equivalent to the change in electric potential energy between the two charges. The formula for potential energy due to two point charges is given by: \[ U = k \frac{q_1 q_2}{r} \]where \(k\) is Coulomb's constant, \(q_1\) and \(q_2\) are the charges, and \(r\) is the distance between the charges.
02

Convert Units

Convert the given coordinates from centimeters to meters to be consistent with SI units.- Charge 1 position: \( (x_1, y_1) = (3.50 \times 10^{-2} \, \text{m}, 0.500 \times 10^{-2} \, \text{m}) \)- Charge 2 position: \( (x_2, y_2) = (-2.00 \times 10^{-2} \, \text{m}, 1.50 \times 10^{-2} \, \text{m}) \)
03

Calculate Distance Between Charges

Calculate the distance \(r\) between the two charges using the distance formula:\[ r = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]Substituting the converted coordinates:\[ r = \sqrt{((-2.00 \times 10^{-2} - 3.50 \times 10^{-2} )^2 + (1.50 \times 10^{-2} - 0.500 \times 10^{-2})^2)} \] \[ = \sqrt{((-5.50 \times 10^{-2})^2 + (1.00 \times 10^{-2})^2)} \] \[ = \sqrt{(0.003025 + 0.0001)} \]\[ = \sqrt{0.003125} \]\[ \approx 0.055884 \text{ meters} \]
04

Substitute Values into Potential Energy Formula

Substitute the values into the potential energy formula:- \( q_1 = 3.00 \times 10^{-6} \text{ C} \)- \( q_2 = -4.00 \times 10^{-6} \text{ C} \)- \( k = 8.99 \times 10^9 \text{ N m}^2/ ext{C}^2 \)- Previously found \( r = 0.055884 \text{ m} \)Substitute into the formula:\[ U = 8.99 \times 10^9 \times \frac{3.00 \times 10^{-6} \times (-4.00 \times 10^{-6})}{0.055884} \]\[ = -8.99 \times 10^9 \times \frac{-12.00 \times 10^{-12}}{0.055884} \]\[ \approx -8.99 \times 10^9 \times (-2.147 \times 10^{-10}) \]\[ \approx -19.2966 \text{ Joules} \]
05

Interpret the Result

The negative sign in the work calculated indicates that work would be required to assemble the charges at the given positions from a state of infinite separation. However, since the question asks for the magnitude of work, we take the absolute value.Therefore, the work required is approximately:\[ 19.2966 \text{ Joules} \]

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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 helps us understand the interaction between two charged particles. It states that the force between two point charges is directly proportional to the product of the magnitude of the charges and inversely proportional to the square of the distance between them. Mathematically, it is expressed as:
  • F = k \frac{q_1 q_2}{r^2}
where:
  • F is the force between the charges.
  • q鈧� and q鈧� are the amounts of the charges.
  • r is the distance separating the charges.
  • k is Coulomb's constant \( (8.99 \times 10^9 \, \text{N m}^2/\text{C}^2) \)
This principle is fundamental because it quantifies how charged objects influence each other. Importantly, the direction of the force is along the line connecting the charges, and it is attractive if the charges are of opposite signs, and repulsive if they are of the same sign.
Point Charges
Point charges are idealized models of particles carrying electric charge. They are imagined as being infinitesimally small, with a definite charge value. In physics problems, especially in basic electrostatics,

we use point charges to simplify calculations. Real-world objects are often replaced with point charges to make the math more manageable.
  • These charges have negligible size.
  • Their effect is only determined by their charge magnitude and position.
  • It simplifies the study of electric fields and forces.
When calculating forces and energies with point charges, we often assume they are in a vacuum. This assumption helps us use simplified formulas like Coulomb's Law without additional factors like medium permittivity interfering.
Work and Energy in Physics
In physics, work and energy are closely related concepts. Work is done when a force moves an object over a distance. In the context of electric forces,

work required to assemble point charges is equivalent to the change in electric potential energy. The potential energy (U) between two charges q鈧� and q鈧� separated by a distance r is given by:
  • \( U = k \frac{q_1 q_2}{r} \)
Here:
  • U is the electric potential energy.
  • If U is negative, the system prefers to be in a nearby configuration, indicating an attractive interaction.
In practical terms, assembling two opposite charges requires work, as they naturally want to be together. Conversely, bringing like charges together demands work against repulsive forces, increasing potential energy in the process.
Coordinate System Conversion
In physics problems, it is often necessary to convert coordinates to use consistent units, especially when working with different measurement systems like meters and centimeters. This often involves converting coordinates to the International System of Units (SI), which is meters in this case.

For example:
  • Charge 1 coordinates in centimeters needed conversion to meters: (3.50 cm, 0.50 cm) became (0.035 m, 0.005 m).
  • Charge 2 went from (-2.00 cm, 1.50 cm) to (-0.020 m, 0.015 m) in the SI unit.
This conversion is crucial as many formulas in physics depend on parameters being in specific units to remain accurate. Ensuring calculations use the correct units avoids errors that can result from unit mismatches.

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Most popular questions from this chapter

What are (a) the charge and (b) the charge density on the surface of a conducting sphere of radius \(0.15 \mathrm{~m}\) whose potential is \(200 \mathrm{~V}\) (with \(V=0\) at infinity)?

A particle of charge \(q\) is fixed at point \(P\), and a second particle of mass \(m\) and the same charge \(q\) is initially held a distance \(r_{1}\) from \(P\). The second particle is then released. Determine its speed when it is a distance \(r_{2}\) from \(P\). Let \(q=3.1 \mu \mathrm{C}, m=20 \mathrm{mg}, r_{1}=\) \(0.90 \mathrm{~mm}\), and \(r_{2}=2.5 \mathrm{~mm}\)

An alpha particle (which has two protons) is sent directly toward a target nucleus containing 92 protons. The alpha particle has an initial kinetic energy of \(0.48 \mathrm{pJ}\). What is the least center-to-center distance the alpha particle will be from the target nucleus, assuming the nucleus does not move?

A proton of kinetic energy \(4.80 \mathrm{MeV}\) travels head-on toward a lead nucleus. Assuming that the proton does not penetrate the nucleus and that the only force between proton and nucleus is the Coulomb force, calculate the smallest center-to-center separation \(d_{p}\) between proton and nucleus when the proton momentarily stops. If the proton were replaced with an alpha particle (which contains two protons) of the same initial kinetic energy, the alpha particle would stop at center-to-center separation \(d_{\alpha^{\circ}}\) (b) What is \(d_{\alpha} / d_{p}\) ?

Initially two electrons are fixed in place with a separation of \(2.00 \mu \mathrm{m} .\) How much work must we do to bring a third electron in from infinity to complete an equilateral triangle?
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