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Chapter 1: Problem 34

\(\cdots\) A charge \(q_{1}=+5.00 \mathrm{nC}\) is placed at the origin of an \(x y\)-coordinate system, and a charge \(q_{2}=-2.00 \mathrm{nC}\) is placed on the positive \(x\)-axis at \(x=4.00 \mathrm{~cm}\). If a third charge \(q_{3}=\) \(+6.00 \mathrm{nC}\) is now placed at the point \(x=4.00 \mathrm{~cm}, y=3.00 \mathrm{~cm}\), find the \(x\) - and \(y\)-components of the total force exerted on this charge by the other two.

Short Answer

Expert verified
The force components are: \( F_x = 8.64 \times 10^{-3} \, \text{N} \), \( F_y = -5.52 \times 10^{-3} \, \text{N} \).

Step by step solution

01

Calculate the Force Between q鈧� and q鈧�

Use Coulomb's Law to find the force between charges at the origin (q鈧�) and at point (4 cm, 3 cm) (q鈧�). The form of Coulomb's Law is \[ F = k \frac{|q_1 q_3|}{r^2} \]where, \( k = 8.99 \times 10^9 \, \text{N m}^2/\text{C}^2 \) is the Coulomb constant, \( q_1 = 5.00 \, \text{nC} = 5.00 \times 10^{-9} \, \text{C} \) and \( q_3 = 6.00 \, \text{nC} = 6.00 \times 10^{-9} \, \text{C}\).First, calculate the distance between the charges using the distance formula:\[ r = \sqrt{(4.00 \times 10^{-2})^2 + (3.00 \times 10^{-2})^2} = 5.00 \times 10^{-2} \, \text{m}\].Now, find the force magnitude:\[ F_{13} = k \frac{|q_1 q_3|}{(5.00 \times 10^{-2})^2} = 8.99 \times 10^{9} \frac{5.00 \times 10^{-9} \times 6.00 \times 10^{-9}}{2.5 \times 10^{-3}} = 1.08 \times 10^{-2} \, \text{N}\].
02

Determine the Direction for F_13 Components

Find the angle \( \theta \) the line makes with the x-axis using trigonometry, where\[ \tan \theta = \frac{y}{x} = \frac{3.00 \times 10^{-2}}{4.00 \times 10^{-2}} \].Thus, \( \theta = \tan^{-1}(0.75) \approx 36.87掳 \).Now, the force components are:\[ F_{13x} = F_{13} \cos(\theta) = 1.08 \times 10^{-2} \times \cos(36.87掳) \approx 8.64 \times 10^{-3} \, \text{N}\]\[ F_{13y} = F_{13} \sin(\theta) = 1.08 \times 10^{-2} \times \sin(36.87掳) \approx 6.48 \times 10^{-3} \, \text{N}\].
03

Calculate the Force Between q鈧� and q鈧�

Use Coulomb's Law for charges at (4 cm, 0) and (4 cm, 3 cm).Distance \( = 3.00 \times 10^{-2} \, \text{m}\) as they have the same x-coordinate.Since \( q_2 = -2.00 \, \text{nC} \), find the force's magnitude:\[ F_{23} = k \frac{|q_2 q_3|}{(3.00 \times 10^{-2})^2} = 8.99 \times 10^{9} \frac{2.00 \times 10^{-9} \times 6.00 \times 10^{-9}}{9.00 \times 10^{-4}} \approx 1.20 \times 10^{-2} \, \text{N}\].
04

Determine the F_23 Direction and Components

Since \( q_2 \) is negative, it attracts \( q_3 \). The force \( F_{23} \) acts towards \( q_2 \) along the y-direction.Thus, the components are:\[ F_{23x} = 0 \]\[ F_{23y} = -1.20 \times 10^{-2} \, \text{N} \] (negative because it's downward on the y-axis as \( q_2 \) is lower than \( q_3 \)).
05

Calculate Total Force on q鈧�

Add the components from forces \( F_{13} \) and \( F_{23} \). Calculate:\[ F_{x,total} = F_{13x} + F_{23x} = 8.64 \times 10^{-3} \, \text{N} + 0 = 8.64 \times 10^{-3} \, \text{N}\].\[ F_{y,total} = F_{13y} + F_{23y} = 6.48 \times 10^{-3} \, \text{N} - 1.20 \times 10^{-2} \, \text{N} = -5.52 \times 10^{-3} \, \text{N}\].

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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 one of the foundational principles in electrostatics, describing how electric charges interact. This law states that the force (F) between two point charges is directly proportional to the product of their magnitudes and inversely proportional to the square of the distance (r^2) between them. The mathematical expression for Coulomb's Law is:
  • \[ F = k \frac{|q_1 q_2|}{r^2} \]
  • Where:
    • \( F \) is the magnitude of the force between the charges.
    • \( k \) is the Coulomb constant \((8.99 \times 10^9 \, \text{N m}^2/\text{C}^2)\).
    • \( q_1 \) and \( q_2 \) are the values of the charges.
    • \( r \) is the distance between the charges.
Coulomb's Law tells us that the force can be either attractive or repulsive, depending on the signs of the charges involved. If the charges are of opposite signs, the force is attractive. If the charges have the same sign, the force is repulsive.
Understanding this law is essential to solving problems involving electric forces in electrostatics.
Electric Forces
Electric forces are the forces of attraction or repulsion due to the electric charges of objects. These forces are a key part of electrostatics and arise from interactions described by Coulomb's Law.
  • For example, in our exercise, multiple forces act on one charge due to the presence of others.
  • The total force on a charge is often the result of the vector sum of all individual electric forces acting on it.
Let鈥檚 look more closely:
  • An electric force can be calculated using each pair of interacting charges.
  • Once calculated, these forces are added using vector addition to determine the overall force experienced by a charge.
This process requires understanding and applying the components of vectors, especially in multi-dimensional situations, as forces typically act in multiple directions simultaneously.
Vector Components
As electricity and magnetism are vector quantities, vector components play a crucial role in analyzing electrostatic problems.
When forces are not aligned with the coordinate axes, they must be broken down into their vector components:
  • The x-component, which runs parallel to the x-axis.
  • The y-component, which runs parallel to the y-axis.
In the exercise, this involves:
  • Using trigonometry, such as sine and cosine functions, based on the angle formed by the force vector relative to the axes.
  • Each force vector is decomposed into two perpendicular components, which are then analyzed separately.
Every force acting on a charge can cause motion either horizontally or vertically, and these sum up to provide the net effect. This is essential when calculating the net force on a charge from multiple interactions.
Charge Interaction
Charge interaction refers to how multiple electric charges interact and affect each other. The nature of these interactions is determined by the sign and magnitude of the charges involved.
For example:
  • Charges of opposite signs attract each other.
  • Charges of the same sign repel each other.
Consideration of charge interaction is crucial in determining:
  • The direction of the electric force experienced by a charge.
  • The magnitude of the resulting net force when multiple charges are present.
In the exercise, the interactions illustrate how each charge experiences forces from the others, and these forces need to be accounted for correctly to find the resultant force. Understanding charge interactions is fundamental for correctly predicting the behavior of charged objects in an electric field.

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Most popular questions from this chapter

Three identical point charges \(q\) are placed at each of three corners of a square of side \(L\). Find the magnitude and direction of the net force on a point charge \(-3 q\) placed (a) at the center of the square and (b) at the vacant corner of the square. In each case, draw a free-body diagram showing the forces exerted on the \(-3 q\) charge by each of the other three charges.

An electron is released from rest in a uniform electric field. The electron accelerates vertically upward, travelling \(4.50 \mathrm{~m}\) in the first \(3.00 \mu s\) after it is released. (a) What are the magnitude and direction of the electric field? (b) Are we justified in ignoring the effects of gravity? Justify your answer quantitatively.

Point charges \(q_{1}=-5 \mathrm{nC}\) and \(q_{2}=+5 \mathrm{nC}\) are separated by \(3 \mathrm{~mm}\), forming an electric dipole. (a) Find the electric dipole moment (magnitude and direction). (b) The charges are in a uniform electric field whose direction makes an angle of \(37^{\circ}\) with the line connecting the charges. What is the magnitude of this field if the torque exerted on the dipole has magnitude \(7.5 \times 10^{-9} \mathrm{~N} \cdot \mathrm{m}\) ?

Two positive point charges \(Q\) are held fixed on the \(x\)-axis at \(x=a\) and \(x=-a\). A third positive point charge \(q\), with mass \(m\), is placed on the \(x\)-axis away from the origin at a coordinate \(x\) such that \(|x| \ll a\). The charge \(q\), which is free to move along the \(x\)-axis, is then released. (a) Find the frequency of oscillation of the charge \(q\). (Himt: Review the definition of simple harmonic motion in Section 14.2. Use the binomial expansion \((1+z)^{n}=1+n z+n(n-1) z^{2} / 2+\cdots\), valid for the case \(|z|<1 .\) ) (b) Suppose instead that the charge \(q\) were placed on the \(y\)-axis at a coordinate \(y\) such that \(|y|

A thin non-conducting ring of radius \(R\) has a linear charge density \(\lambda=\lambda_{0} \cos \theta\), where \(\lambda_{0}\) is the value of \(\lambda\) at \(\theta=0\). Find net electric dipole moment for this charge distribution.
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