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An electrically neutral model airplane is flying in a horizontal circle on a 3.0 -m guideline, which is nearly parallel to the ground. The line breaks when the kinetic energy of the plane is \(50.0 \mathrm{~J}\). Reconsider the same situation, except that now there is a point charge of \(+q\) on the plane and a point charge of \(-q\) at the other end of the guideline. In this case, the line breaks when the kinetic energy of the plane is \(51.8 \mathrm{~J}\). Find the magnitude of the charges.

Short Answer

Expert verified
The magnitude of each charge is approximately \(3.51 \times 10^{-5} \, \text{C}\).

Step by step solution

01

Calculate the Force in Both Cases

The line breaks when the kinetic energy of the plane is at specific values, indicating the tension equals the centripetal force. First, for the neutral plane: \( F_{1} = \frac{KE_{1} \cdot 2}{r} = \frac{50 \, \text{J} \cdot 2}{3.0 \, \text{m}} = 33.3 \, \text{N}\). For the charged plane: \( F_{2} = \frac{KE_{2} \cdot 2}{r} = \frac{51.8 \, \text{J} \cdot 2}{3.0 \, \text{m}} = 34.53 \, \text{N}\).
02

Express the Coulomb's Law Force

The difference in force between the two cases is due to the electric force between charges. The formula for the electric force is given by Coulomb's Law: \( F_e = \frac{k \cdot q^2}{r^2} \). Here, \( r = 3.0 \, \text{m} \) and \( k = 8.99 \times 10^9 \, \text{N m}^2/\text{C}^2 \).
03

Determine the Difference in Force

Set the electric force equal to the difference between the breaking tensions: \( F_e = F_{2} - F_{1} = 34.53 \, \text{N} - 33.3 \, \text{N} = 1.23 \, \text{N}\).
04

Solve for the Charge Magnitude

Using the equation \( F_e = \frac{k \cdot q^2}{r^2} \), solve for \( q \): \[ q^2 = \frac{1.23 \, \text{N} \times (3.0 \, \text{m})^2}{8.99 \times 10^9 \, \text{N m}^2/\text{C}^2} \] \[ q^2 = 1.23 \times 9 / 8.99 \times 10^9 \; \Rightarrow \; q^2 = 1.107 / 8.99 \times 10^9 \]\[ q \approx \sqrt{1.23 \times 9 / 8.99 \times 10^9} \approx \sqrt{1.107 / 8.99 \times 10^9} \] \( q \approx 3.51 \times 10^{-5} \; \text{C} \).
05

Final Result Verification

Double-check the calculation of \( q \) to ensure no computational errors by redoing the multiplication and division steps. Confirm \( q = 3.51 \times 10^{-5} \, \text{C} \), consistent with the derived equations.

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Key Concepts

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

Electromagnetism
Electromagnetism is the branch of physics that studies electric charges, electric and magnetic fields, and their interactions. It's a fundamental aspect of physics that explains how electrically charged particles interact with each other and with magnetic fields. This concept combines two main forces: electricity and magnetism, which are typically studied together because they are deeply interconnected.
Electromagnetic forces are essential for understanding how motors work, how light travels, and even how atoms and molecules are held together. They exert powerful effects over long distances.
  • Electric charges generate electric fields.
  • Moving charges produce magnetic fields.
  • Changes in magnetic fields can induce electric currents, according to Faraday's Law.

In the context of the exercise, the aircraft maintains stability under electromagnetic forces due to its charge, demonstrating the broad applicability of electromagnetic theory.
Coulomb's Law
Coulomb's Law is a crucial principle in electromagnetism, describing the force between two charged objects. The formula is given by:\[ F_e = \frac{k \cdot q_1 \cdot q_2}{r^2} \]where \( F_e \) is the electric force, \( k \) is Coulomb's constant \((8.99 \times 10^9 \, \text{Nm}^2/\text{C}^2)\), \( q_1 \) and \( q_2 \) are the charges, and \( r \) is the distance between them.
This force is inversely proportional to the square of the distance, which means that as charges move farther apart, the force diminishes quickly.
  • Uses point charges and assumes they are at rest or moving very slowly.
  • Explores both attractive and repulsive forces depending on the sign of the charges.
In the exercise, using Coulomb's Law allowed us to calculate the additional force needed to break the guideline due to charged particles.
Centripetal Force
Centripetal force is essential for any object moving in a circular path, acting perpendicular to the velocity of the object and keeping it on its path. It is directed toward the center of the circle around which the object rotates.The formula is expressed as:\[ F_c = \frac{mv^2}{r} \]where \( m \) is the mass of the object, \( v \) is its velocity, and \( r \) is the radius of the circle.
  • This force does not exist by itself but results from other forces (like tension, gravity, etc.).
  • It's crucial to note that without this force, an object would continue in a straight line (Newton's First Law of Motion).
In the exercise demonstrating the breakage of the guideline, the tension providing centripetal force became crucial when external charges introduced additional force.
Kinetic Energy
Kinetic energy is the energy possessed by an object due to its motion, directly proportional to its mass and the square of its velocity. The formula for kinetic energy \( KE \) is:\[ KE = \frac{1}{2} mv^2 \]where \( m \) is the mass and \( v \) is the velocity of the object.
In any closed system, kinetic energy plays a significant role in energy conservation, often compared or transferred into potential energy.
  • A key parameter in collision and explosion analysis.
  • Vital in understanding work-energy principles.

The exercise outlines how the kinetic energy increased when electric charges were added, leading to the guideline breaking, which indicates how dynamic motion interacts with typical forces.

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Most popular questions from this chapter

There are four charges, each with a magnitude of \(2.0 \mu \mathrm{C}\). Two are positive and two are negative. The charges are fixed to the corners of a 0.30 -m square, one to a corner, in such a way that the net force on any charge is directed toward the center of the square. Find the magnitude of the net electrostatic force experienced by any charge.

A charge \(+q\) is located at the origin, while an identical charge is located on the \(x\) axis at \(x=+0.50 \mathrm{~m}\). A third charge of \(+2 q\) is located on the \(x\) axis at such a place that the net electrostatic force on the charge at the origin doubles, its direction remaining unchanged. Where should the third charge be located?

A rectangle has a length of \(2 d\) and a height of \(d\). Each of the following three charges is located at a corner of the rectangle: \(+q_{1}\) (upper left corner), \(+q_{2}\) (lower right corner), and \(-q\) (lower left corner). The net electric field at the (empty) upper right corner is zero. Find the magnitudes of \(q_{1}\) and \(q_{2}\). Express your an swers in terms of \(q\).

Interactive LearningWare 18.1 at offers some perspective on this problem. Two tiny spheres have the same mass and carry charges of the same magnitude. The mass of each sphere is \(2.0 \times 10^{-6} \mathrm{~kg} .\) The gravitational force that each sphere exerts on the other is balanced by the electric force. (a) What algebraic signs can the charges have? (b) Determine the charge magnitude.

Consult Concept Simulation 18.1 at for insight into this problem. Three charges are fixed to an \(x, y\) coordinate system. A charge of \(+18 \mu \mathrm{C}\) is on the \(y\) axis at \(y=+3.0 \mathrm{~m}\). A charge of \(-12 \mu \mathrm{C}\) is at the origin. Last, a charge of \(+45 \mu \mathrm{C}\) is on the \(x\) axis at \(x=+3.0 \mathrm{~m}\). Determine the magnitude and direction of the net electrostatic force on the charge at \(x=+3.0 \mathrm{~m}\). Specify the direction relative to the \(-x\) axis.

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