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Three point charges have equal magnitudes, two being positive and one negative. These charges are fixed to the corners of an equilateral triangle, as the drawing shows. The magnitude of each of the charges is \(5.0 \mu \mathrm{C},\) and the lengths of the sides of the triangle are \(3.0 \mathrm{cm} .\) Calculate the magnitude of the net force that each charge experiences.

Short Answer

Expert verified
Each charge experiences a net force of 2.49 N.

Step by step solution

01

Identify the Forces on a Charge

Since it's an equilateral triangle, each angle is 60°. Consider one of the positive charges and note that the net force on it will be affected by both the other positive charge and the negative charge. Let's denote the charges as \( Q_1, Q_2, \) and \( Q_3 \), where \( Q_1 \) and \( Q_2 \) are positive, and \( Q_3 \) is negative. We'll examine the forces on \( Q_1 \).
02

Force between Q1 and Q2

Using Coulomb's Law, the force between \( Q_1 \) and \( Q_2 \) is repulsive as they are both positive. The formula is: \[ F_{12} = k \frac{|Q_1 Q_2|}{r^2} \]Substituting the known values: \[ F_{12} = (8.99 \times 10^9) \frac{(5.0 \times 10^{-6})^2}{(0.03)^2} = 2.49 N \]
03

Force between Q1 and Q3

The force between \( Q_1 \) and \( Q_3 \) is attractive because of the opposite charges. Again using Coulomb's Law:\[ F_{13} = k \frac{|Q_1 Q_3|}{r^2} \]Substitute the known values: \[ F_{13} = (8.99 \times 10^9) \frac{(5.0 \times 10^{-6})^2}{(0.03)^2} = 2.49 N \]
04

Resolve Forces into Components

Since the net force on \( Q_1 \) due to \( Q_2 \) is along the line joining \( Q_1 \) and \( Q_2 \), and similarly for \( Q_3 \), we need to resolve them into x and y components. The angle between each force and the horizontal axis is 30°.For the x-component:\[ F_{x1} = F_{12} \cos(30°) - F_{13} \cos(30°) = 0 \]For the y-component:\[ F_{y1} = F_{12} \sin(30°) + F_{13} \sin(30°) = 2 \times 2.49 \times \frac{1}{2} = 2.49 N \]
05

Calculate the Net Force

The net force will be solely in the vertical direction (y-direction) since the horizontal components cancel each other out:\[ F_{net} = 2.49 N \]
06

Repeat for Other Charges

The symmetry of the problem suggests that the magnitude of the net force experienced by the other charges (\( Q_2 \) and \( Q_3 \)) will be the same due to the equal magnitudes and similar positions within the equilateral triangle.

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Key Concepts

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

Electrostatics
Electrostatics is the study of electric charges at rest. It examines the forces, fields, and potential energies associated with stationary or slow-moving charges. One of the key principles in electrostatics is Coulomb's Law. This law describes the force between two point charges. According to Coulomb's Law, the force (\( F \)) between two charges is directly proportional to the product of their magnitudes and inversely proportional to the square of the distance between them. In mathematical terms: \[ F = k \frac{|Q_1 Q_2|}{r^2} \] where \( k \) is Coulomb's constant. For practical calculations in electrostatics, especially when charges are arranged in geometric patterns such as triangles, understanding the vector nature of forces is crucial. Each charge can both repel and attract depending on the nature of surrounding charges (positive vs. negative), creating a net electrostatic force that must be resolved into its component parts.
Net Force Calculation
In problems involving multiple charges, calculating the net force on a charge involves several steps. The first step is determining the force between each pair of charges using Coulomb's Law. Once these forces are known, they must be resolved into their respective components. This involves breaking down each force into its x and y components using trigonometric functions based on the geometry of the problem. For example, in an equilateral triangle, symmetry allows certain simplifications:
  • All sides and angles are equal, which simplifies the calculation of distances and angles.
  • The forces can be resolved into perpendicular components where symmetry can lead to cancellation of certain components.
  • The components along the axes are found using sine and cosine of the relevant angle.
After resolving the components, you calculate the net force by summing the vectors of all the components. The horizontal components often cancel each other out in systems with symmetry, leaving a simplified net force in one direction.
Equilateral Triangle Charges
Placing charges at the vertices of an equilateral triangle provides an excellent example of symmetry in electrostatics. An equilateral triangle has equal side lengths and angles of 60°. This symmetry can simplify the calculation of net forces because similar distance and angle relations repeat for each charge. Let's break down the scenario:
  • Each charge in an equilateral triangle experiences forces from the other two charges.
  • The forces between charges that share the same sign (both positive or both negative) are repulsive.
  • The forces between charges with opposite signs are attractive.
  • Due to the 60° angles, components of these forces can be easily calculated, often leading to exact cancellations in the horizontal direction.
In such configurations, the net force often points entirely in the vertical direction due to the balanced arrangement and the symmetry of forces along the triangle's axes. The net force experienced by each charge is typically only influenced by the vertical components when horizontal forces cancel out due to the symmetry.

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Most popular questions from this chapter

Two identical small insulating balls are suspended by separate \(0.25-\mathrm{m}\) threads that are attached to a common point on the ceiling. Each ball has a mass of \(8.0 \times 10^{-4} \mathrm{kg} .\) Initially the balls are uncharged and hang straight down. They are then given identical positive charges and, as a result, spread apart with an angle of \(36^{\circ}\) between the threads. Determine (a) the charge on each ball and (b) the tension in the threads.

ssm Two particles, with identical positive charges and a separation of \(2.60 \times 10^{-2} \mathrm{m},\) are released from rest. Immediately after the release, particle 1 has an acceleration \(\overrightarrow{\mathbf{a}}_{1}\) whose magnitude is \(4.60 \times 10^{3} \mathrm{m} / \mathrm{s}^{2},\) while particle 2 has an acceleration \(\overrightarrow{\mathbf{a}}_{2}\) whose magnitude is \(8.50 \times 10^{3} \mathrm{m} / \mathrm{s}^{2}\). Particle 1 has a mass of \(6.00 \times 10^{-6} \mathrm{kg} .\) Find (a) the charge on each particle and (b) the mass of particle 2.

There are four charges, each with a magnitude of \(2.0 \mu\) C. Two are positive and two are negative. The charges are fixed to the corners of a \(0.30-\mathrm{m}\) 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 small drop of water is suspended motionless in air by a uniform electric field that is directed upward and has a magnitude of \(8480 \mathrm{N} / \mathrm{C}\). The mass of the water drop is \(3.50 \times 10^{-9} \mathrm{kg} .\) (a) Is the excess charge on the water drop positive or negative? Why? (b) How many excess electrons or protons reside on the drop?

ssm Two very small spheres are initially neutral and separated by a distance of \(0.50 \mathrm{m}\). Suppose that \(3.0 \times 10^{13}\) electrons are removed from one sphere and placed on the other. (a) What is the magnitude of the electrostatic force that acts on each sphere? (b) Is the force attractive or repulsive? Why?

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