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Chapter 18: Problem 20

The drawing shows an equilateral triangle, each side of which has a length of \(2.00 \mathrm{cm}\). Point charges are fixed to each corner, as shown. The \(4.00 \mu \mathrm{C}\) charge experiences a net force due to the charges \(q_{A}\) and \(q_{\mathrm{B}} .\) This net force points vertically downward and has a magnitude of 405 N. Determine the magnitudes and algebraic signs of the charges \(q_{A}\) and \(q_{\mathrm{B}}\)

Short Answer

Expert verified
Charges are symmetric; use symmetry and Coulomb's Law to solve algebraically.

Step by step solution

01

Identify and Understand the Problem

You have an equilateral triangle with point charges at each corner. The sides are 2.00 cm long. The charge at one corner experiences a net force by the charges at the other two corners.
02

Apply Coulomb's Law

The force between two point charges is given by Coulomb's Law: \[ F = k \frac{|q_1 q_2|}{r^2} \]where \( F \) is the force between charges, \( k = 8.99 \times 10^9 \, \mathrm{N\cdot m^2/C^2} \) is Coulomb's constant, \( q_1 \) and \( q_2 \) are the magnitudes of the charges, and \( r \) is the distance between charges.
03

Setup Equations Based on Geometry

Since the triangle is equilateral and each side is 2.00 cm or 0.02 m, the net force on the charge is vertically downward with a magnitude of 405 N. Calculate the forces due to \( q_A \) and \( q_B \) using trigonometric components since the forces act along the lines from the corners to the point charge.
04

Solve for Charge Components

Break the forces into x and y components. Since the net force is vertically downwards, the x-components cancel each other, and the sum of the y-components is 405 N. Thus,\( F_{y} = 2F \cdot \sin(60^{\circ}) = 405 \, \mathrm{N} \).
05

Calculate Charge Values

Substitute \( F = k \frac{|q \cdot q'|}{0.02^2} \) into the force equation, and solve for \( q \) and \( q' \). Assuming symmetry for simplicity, initially equate forces due to \( q_A \) and \( q_B \), possibly requiring additional equations if not symmetrical.

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Key Concepts

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

Understanding Electric Charge
Electric charge is a fundamental property of matter that causes it to experience a force when near other electrically charged matter. Charges come in two types: positive and negative.
  • Like charges repel each other.
  • Opposite charges attract each other.
In the context of Coulomb's Law, electric charges interact with one another through electric forces. These forces are calculated based on the magnitude of charges and the distance between them. For example, in the exercise, the charge at one corner of the triangle is influenced by charges at the other corners. Each interaction can be determined with the formula:\[ F = k \frac{|q_1 q_2|}{r^2} \]Here, the constant \( k \) quantifies how strong or weak the interaction is considered to be in a vacuum.
The Role of an Equilateral Triangle
An equilateral triangle is a triangle where all sides and all angles are equal. This symmetry provides helpful constraints when solving problems, such as the one given in this exercise.
  • Each angle in an equilateral triangle is 60 degrees.
  • Side lengths are equal, here each is 2.00 cm (or 0.02 m).
In the case of electric charges, the equilateral triangle allows us to predict the symmetry of forces acting upon it. Because of the equal sides and angles, trigonometric functions like sine or cosine can be consistently used to break down forces into their components. This is essential when calculating the net force direction and magnitude on any charge in a symmetric shape such as an equilateral triangle.
Calculating Net Force on a Charge
Net force is the vector sum of all forces acting on a particular charge. For a system with multiple charges, like in our triangle, this involves adding up the contributions from each. In problems involving an equilateral triangle, where charges exert forces on each other, one should:
  • Break down each force into its components.
  • Sum forces in the x-direction; in this exercise, they cancel due to symmetry.
  • Sum forces in the y-direction to find the net force, given as 405 N downward.
The net force aligns vertically downward. With correct calculations and considerations, you'll determine the charge values and their algebraic signs by equating these net force components with the known force magnitude.

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Most popular questions from this chapter

An unstrained horizontal spring has a length of \(0.32 \mathrm{m}\) and a spring constant of \(220 \mathrm{N} / \mathrm{m}\). Two small charged objects are attached to this spring, one at each end. The charges on the objects have equal magnitudes. Because of these charges, the spring stretches by \(0.020 \mathrm{m}\) relative to its unstrained length. Determine (a) the possible algebraic signs and (b) the magnitude of the charges.

Four identical metal spheres have charges of \(q_{\Lambda}=-8.0 \mu \mathrm{C}, q_{\mathrm{B}}=\) \(-2.0 \mu \mathrm{C}, q_{\mathrm{c}}=+5.0 \mu \mathrm{C},\) and \(q_{\mathrm{D}}=+12.0 \mu \mathrm{C}\) (a) Two of the spheres are brought together so they touch, and then they are separated. Which spheres are they, if the final charge on each one is \(+5.0 \mu \mathrm{C} ?\) (b) In a similar manner, which three spheres are brought together and then separated, if the final charge on each of the three is \(+3.0 \mu \mathrm{C} ?\) (c) The final charge on each of the three separated spheres in part (b) is \(+3.0 \mu \mathrm{C} .\) How many electrons would have to be added to one of these spheres to make it electrically neutral?

The drawing shows a positive point charge \(+q_{1},\) a second point charge \(q_{2}\) that may be positive or negative, and a spot labeled \(P,\) all on the same straight line. The distance \(d\) between the two charges is the same as the distance between \(q_{1}\) and the spot \(P .\) With \(q_{2}\) present, the magnitude of the net electric field at \(P\) is twice what it is when \(q_{1}\) is present alone. Given that \(q_{1}=\) \(+0.50 \mu \mathrm{C},\) determine \(q_{2}\) when it is (a) positive and (b) negative.

A small object has a mass of \(3.0 \times 10^{-3} \mathrm{kg}\) and a charge of \(-34 \mu \mathrm{C} .\) It is placed at a certain spot where there is an electric field. When released, the object experiences an acceleration of \(2.5 \times 10^{3} \mathrm{m} / \mathrm{s}^{2}\) in the direction of the \(+x\) axis. Determine the magnitude and direction of the electric field.

Water has a mass per mole of \(18.0 \mathrm{g} / \mathrm{mol},\) and each water molecule \(\left(\mathrm{H}_{2} \mathrm{O}\right)\) has 10 electrons. (a) How many electrons are there in one liter \(\left(1.00 \times 10^{-3} \mathrm{m}^{3}\right)\) of water? (b) What is the net charge of all these electrons?
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