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Chapter 1: Problem 33

Four identical charges \(Q\) are placed at the corners of a square of side \(L\). (a) In a free-body diagram, show all of the forces that act on one of the charges. (b) Find the magnitude and direction of the total force exerted on one charge by the other three charges.

Short Answer

Expert verified
Identify all forces, calculate individual forces using Coulomb's law, resolve diagonal forces, sum components, and find total force magnitude and direction.

Step by step solution

01

Understand the Forces on a Charge

Each charge is repelled by the other three due to their like charges. We will consider one charge and identify the forces due to the other three. Let鈥檚 call the charge of interest at the origin as charge A, and the others as B, C, and D positioned at the respective corners of the square.
02

Represent Forces in a Free-Body Diagram

Place charge A and indicate the forces: The repulsion from charge B will act along the horizontal (+x direction), from charge C it will act vertically (+y direction), and from charge D (diagonally across the square) it will act along the diagonal.
03

Calculate the Force Due to Adjacent Charges

The force between two point charges using Coulomb's Law is given by \[ F = k \cdot \frac{Q^2}{d^2} \]where \(k\) is Coulomb's constant, \(Q\) is the charge, and \(d\) is the distance. For adjacent charges (\(B\) and \(C\)), \(d = L\). Therefore, \[ F_{B} = F_{C} = k \cdot \frac{Q^2}{L^2} \]
04

Calculate the Force Due to the Diagonal Charge

The charge diagonally across (\(D\)) is at a distance \( \sqrt{2}L \). Hence, \[ F_{D} = k \cdot \frac{Q^2}{(\sqrt{2}L)^2} = k \cdot \frac{Q^2}{2L^2} \]
05

Resolve Diagonal Force into Components

The force from \(D\) needs to be divided into its components along the x and y axes. Since the angle with respect to the x-axis will be \(45^\circ\), the components are:\[ F_{Dx} = F_{D} \cdot \cos(45^\circ) = \frac{k \cdot Q^2}{2L^2} \cdot \frac{1}{\sqrt{2}} = \frac{k \cdot Q^2}{2\sqrt{2}L^2} \]\[ F_{Dy} = F_{D} \cdot \sin(45^\circ) = \frac{k \cdot Q^2}{2L^2} \cdot \frac{1}{\sqrt{2}} = \frac{k \cdot Q^2}{2\sqrt{2}L^2} \]
06

Calculate the Total Force Magnitude

Add the components of the forces in both x and y directions:For x-direction, \[ F_{x} = F_{B} + F_{Dx} = \frac{k \cdot Q^2}{L^2} + \frac{k \cdot Q^2}{2\sqrt{2}L^2} \]For y-direction, \[ F_{y} = F_{C} + F_{Dy} = \frac{k \cdot Q^2}{L^2} + \frac{k \cdot Q^2}{2\sqrt{2}L^2} \]The total force magnitude \( F \) is given by: \[ F = \sqrt{F_{x}^2 + F_{y}^2} \]
07

Determine the Direction of the Total Force

The direction \( \theta \) can be determined from the ratio of \( F_y \) to \( F_x \):\[ \tan \theta = \frac{F_{y}}{F_{x}} \]This angle gives the direction of the net force with respect to the x-axis.

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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 a fundamental principle in electrostatics that calculates the force between two point charges. It was formulated by Charles-Augustin de Coulomb in the 18th century. The law states that the electric force (\( F \)}) between two charges is proportional to the product of the charges' magnitudes and inversely proportional to the square of the distance between them. Mathematically, it is expressed as:\[ F = k \cdot \frac{Q_1 \cdot Q_2}{r^2} \]where:
  • \( k \) is the Coulomb's constant, approximately equal to \( 8.99 \times 10^9 \text{ Nm}^2/\text{C}^2 \)
  • \( Q_1 \) and \( Q_2 \) are the point charges
  • \( r \) is the distance separating the charges
This law helps us understand how charged particles interact with each other, governing both the attraction and repulsion forces based on their signs. In problems involving multiple charges, such as the one in the exercise, Coulomb's Law is applied individually to each pair of charges to find the net force on a particular charge.
Electric Force
Electric force is the force exerted between charged objects. In the case of point charges, it is calculated using Coulomb's Law, as previously discussed. The electric force is a vector, meaning it has both magnitude and direction.

In electrostatic problems, such as when four charges are placed at the corners of a square, the electric force on a particular charge is due to the influence of the other three charges. The forces from adjacent charges typically act along the lines connecting them, causing repulsion when charges are like-signed. For a charge, this means calculating the force from each of the other three individually and then adding these vectorially to find the total electric force experienced by the charge in question.
  • The force from charges directly across an axis often aligns with that axis.
  • Forces from diagonally positioned charges have both x and y components, requiring vector resolution.
Vector Components
Understanding vector components is crucial in analyzing forces in two dimensions. When dealing with electric forces in a square configuration, each force from a different charge can be broken down into two perpendicular components: x and y.

For example, if a force acts diagonally, this force can be split into horizontal (x-axis) and vertical (y-axis) components using trigonometric functions. For geometric angles like 45 degrees, which are common in square setups:
  • The x-component of the force (\( F_x \)) is calculated as \( F \cdot \cos(45^\circ) \).
  • The y-component of the force (\( F_y \)) is calculated as \( F \cdot \sin(45^\circ) \).
These components are essential for determining the cumulative effect of all forces in the square, allowing us to compute the total force in each direction. Ultimately, they help in determining the overall net force acting on a charge.
Free-Body Diagram
A free-body diagram is a useful tool in physics to visualize forces acting on a single object. For electrostatic problems involving multiple charges, the diagram maps how each charge interacts with a focal charge.

In this exercise, we focus on one charge (let's call it charge A) within a square of four like charges. The free-body diagram shows all the forces exerted on charge A by the other three. These forces include:
  • A horizontal force along the x-direction from the adjacent charge to the side.
  • A vertical force along the y-direction from the adjacent charge above or below.
  • A diagonal force from the charge positioned diagonally opposite, which will need component resolution into x and y parts.
A properly constructed free-body diagram provides a clear visual representation, making it easier to solve for the net force by allowing straightforward vector analysis.

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Most popular questions from this chapter

A \(+2.00-n C\) point charge is at the origin, and a second \(-5.00-n C\) point charge is on the \(x\)-axis at \(x=0.800 \mathrm{~m} .\) (a) Find the electric field (magnitude and direction) at each of the following points on the \(x\)-axis: (i) \(x=0.200 \mathrm{~m}\); (ii) \(x=-0.200 \mathrm{~m}\). (b) Find the net electric force that the two charges would exert on an electron placed at each point in part (a).

A proton is projected into a uniform electric field that points vertically upward and has magnitude \(E\). The initial velocity of the proton has a magnitude \(v_{0}\) and is directed at an angle \(\alpha\) below the horizontal. (a) Find the maximum distance \(h_{\max }\) that the proton descends vertically below its initial elevation. You can ignore gravitational forces. (b) After what horizontal distance \(d\) does the proton return to its original elevation? (c) Sketch the trajectory of the proton. (d) Find the numerical values of \(h_{\max }\) and \(d\) if \(E=500 \mathrm{~N} / \mathrm{C}, v_{0}=4.00 \times 10^{5} \mathrm{~m} / \mathrm{s}\), and \(\alpha=30.0^{\circ}\)

Positive charge \(+Q\) is distributed uniformly along the \(+x\)-axis from \(x=0\) to \(x=a\). Negative charge \(-Q\) is distributed uniformly along the \(-x\)-axis from \(x=0\) to \(x=-a\). (a) A positive point charge \(q\) lies on the positive \(y\)-axis, a distance \(y\) from the origin. Find the force (magnitude and direction) that the positive and negative charge distributions together exert on \(q\). Show that this force is proportional to \(y^{-3}\) for \(y \gg a\). (b) Suppose instead that the positive point charge \(q\) lies on the positive \(x\)-axis, a distance \(x>a\) from the origin. Find the force (magnitude and direction) that the charge distribution exerts on \(q\). Show that this force is proportional to \(x^{-3}\) for \(x \gg a\).

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 \(7.5 \times 10^{-9} \mathrm{~N} \cdot \mathrm{m}\) ?

A very long, straight wire has charge per unit length \(1.50 \times 10^{-10} \mathrm{C} / \mathrm{m}\). At what distance from the wire is the electricfield magnitude equal to \(2.50 \mathrm{~N} / \mathrm{C} ?\)
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