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Chapter 18: Problem 63

Two charges are located along the \(x\) axis: \(q_{1}=+6.0 \mu \mathrm{C}\) at \(x_{1}=+4.0 \mathrm{~cm}\), and \(q_{2}=+6.0 \mu \mathrm{C}\) at \(x_{2}=-4.0 \mathrm{~cm}\). Two other charges are located on the \(y\) axis: \(q_{3}=+3.0 \mu \mathrm{C}\) at \(y_{3}=+5.0 \mathrm{~cm}\), and \(q_{4}=-8.0 \mu \mathrm{C}\) at \(y_{4}=+7.0 \mathrm{~cm} .\) Find the net electric field (magnitude and direction) at the origin.

Short Answer

Expert verified
The net electric field at the origin is \(3.9 \times 10^{6}\) N/C upward.

Step by step solution

01

Determine Distance from Charges to Origin

The origin is located at \((x, y) = (0, 0)\). The distances from each charge to the origin can be calculated using thecoordinates of the charges:\[\text{For } q_1: \sqrt{(x_1)^2} = \sqrt{4.0^2} = 4.0 \text{ cm (0.04 m)},\]\[\text{For } q_2: \sqrt{(x_2)^2} = \sqrt{(-4.0)^2} = 4.0 \text{ cm (0.04 m)},\]\[\text{For } q_3: \sqrt{(y_3)^2} = \sqrt{5.0^2} = 5.0 \text{ cm (0.05 m)},\]\[\text{For } q_4: \sqrt{(y_4)^2} = \sqrt{7.0^2} = 7.0 \text{ cm (0.07 m)}.\]
02

Calculate Electric Field Contribution from Each Charge

The electric field due to a point charge is given by:\[ E = \frac{k |q|}{r^2} \]where \( k = 8.99 \times 10^9 \text{ N m}^2/\text{C}^2 \) is Coulomb's constant.**For \( q_1 \):**\(E_1 = \frac{8.99 \times 10^9 \times 6.0 \times 10^{-6}}{(.04)^2} = 3.37 \times 10^7 \text{ N/C}\), directed left due to being on the \( x \)-axis.**For \( q_2 \):**\(E_2 = \frac{8.99 \times 10^9 \times 6.0 \times 10^{-6}}{(.04)^2} = 3.37 \times 10^7 \text{ N/C}\), directed right due to being on the \( x \)-axis.**For \( q_3 \):**\(E_3 = \frac{8.99 \times 10^9 \times 3.0 \times 10^{-6}}{(.05)^2} = 1.08 \times 10^7 \text{ N/C}\), directed down due to being on the \( y \)-axis.**For \( q_4 \):**\(E_4 = \frac{8.99 \times 10^9 \times 8.0 \times 10^{-6}}{(.07)^2} = 1.47 \times 10^7 \text{ N/C}\), directed up due to being negatively charged.
03

Determine the Net Electric Field Components

The net electric field at the origin is found by summing the vector components of the electric fields due to each charge.**Horizontal Component (\(E_x\)) from \( q_1 \) and \( q_2\):**Both are equal but opposite, so they cancel:\( E_{x~\text{net}} = 3.37 \times 10^7 - 3.37 \times 10^7 = 0 \text{ N/C} \).**Vertical Component (\(E_y\)) from \( q_3 \) and \( q_4\):**\( E_{y~\text{net}} = 1.47 \times 10^7 - 1.08 \times 10^7 = 0.39 \times 10^7 \text{ N/C} \), directed upwards.
04

Calculate the Net Electric Field Magnitude and Direction

Since the horizontal component is zero, the magnitude of the net electric field is simply the absolute value of the vertical component:\(E_{\text{net}} = |E_{y~\text{net}}| = 0.39 \times 10^7 = 3.9 \times 10^6 \text{ N/C}\)The direction is upwards along the \( y \)-axis as calculated.

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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 that describes the electric force between two charged objects. It states that the force (\( F \)) between two point charges is directly proportional to the product of the absolute values of the charges (\( q_1 \) and \( q_2 \)), and inversely proportional to the square of the distance (\( r \)) between them. Mathematically, it is expressed as:\[ F = \frac{k |q_1 q_2|}{r^2} \]where \( k = 8.99 \times 10^9 \, \text{N m}^2/\text{C}^2 \) is Coulomb's constant. Coulomb's law is crucial for calculating the electric field caused by point charges, like those in the exercise. Understanding this law helps us determine how charged objects interact, influencing their behavior within an electric field.
Electric charge
Electric charge is a fundamental property of matter, responsible for electric forces and fields. It comes in two types: positive and negative. Like charges repel, while unlike charges attract. This behavior is foundational in understanding many physical phenomena.Electric charge is measured in coulombs (C). A charge can be described as either a point charge, which focuses on a single point in space, or a distributed charge over a larger region. The charges in the exercise are point charges with specific magnitudes, such as \( q_1 = +6.0 \, \mu \text{C}\). Charges are often found in microcoulombs (\( \mu\text{C} \)), where \( 1 \mu\text{C} = 10^{-6} \text{C} \), indicating a smaller quantity of charge used in practical applications.
Vector components
Vector components are crucial for analyzing forces and fields in multiple dimensions. An electric field, which is a vector quantity, possesses both magnitude and direction. To compute the net field at a point, we must consider its individual vector components along designated axes (e.g., x-axis and y-axis).In the exercise, charges \( q_1 \) and \( q_2 \) lie along the x-axis, contributing horizontal components to the electric field. Similarly, \( q_3 \) and \( q_4 \) influence the field vertically along the y-axis. By decomposing these fields into vector components, we easily calculate the net electric field:
  • Horizontal components can cancel each other if equal and opposite.
  • Vertical components are summed to find the resultant net force direction.
Understanding vector components helps solve physics problems involving directions, such as finding the net force or field at a particular point.
Point charge
A point charge refers to an electric charge that is considered to exist at a single point in space. This idealized concept simplifies calculations of electric fields, as the field can be treated as originating from one concentrated point. In practice, real charged objects have spatial dimensions, but treating them as point charges is useful when those dimensions are negligible compared to the distance involved. In the exercise, each given charge is treated as a point charge, which makes it easier to apply formulas like Coulomb's law directly. The use of point charges allows for clean, straightforward calculations of interactions, predicting how these charges influence others and the resulting electric field established at any given point, such as the origin in this exercise.

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Most popular questions from this chapter

In a vacuum, two particles have charges of \(q_{1}\) and \(q_{2}\), where \(q_{1}=+3.5 \mu \mathrm{C}\). They are separated by a distance of \(0.26 \mathrm{~m}\), and particle 1 experiences an attractive force of \(3.4 \mathrm{~N}\). What is \(q_{2}\) (magnitude and sign)?

Two parallel plate capacitors have circular plates. The magnitude of the charge on these plates is the same. However, the electric field between the plates of the first capacitor is \(2.2 \times 10^{5} \mathrm{~N} / \mathrm{C},\) while the field within the second capacitor is \(3.8 \times 10^{5} \mathrm{~N} / \mathrm{C}\). Determine the ratio \(r_{2} / r_{1}\) of the plate radius for the second capacitor to the plate radius for the first capacitor.

A cube is located with one corner at the origin of an \(x, y, z\) coordinate system. One of the cube's faces lies in the \(x, y\) plane, another in the \(y, z\) plane, and another in the \(x, z\) plane. In other words, the cube is in the first octant of the coordinate system. The edges of the cube are \(0.20 \mathrm{~m}\) long. A uniform electric field is parallel to the \(x, y\) plane and points in the direction of the \(+y\) axis. The magnitude of the field is \(1500 \mathrm{~N} / \mathrm{C}\). (a) Find the electric flux through each of the six faces of the cube. (b) Add the six values obtained in part (a) to show that the electric flux through the cubical surface is zero, as Gauss' law predicts, since there is no net charge within the cube.

Two charges attract each other with a force of \(1.5 \mathrm{~N}\). What will be the force if the distance between them is reduced to one-ninth of its original value?

A long, thin, straight wire of length \(L\) has a positive charge \(Q\) distributed uniformly along it. Use Gauss' law to show that the electric field created by this wire at a radial distance \(r\) has a magnitude of \(E=\lambda /\left(2 \pi \epsilon_{0} r\right)\), where \(\lambda=Q / L .\) (Hint: For a Gaussian surface, use a cylinder aligned with its axis along the wire and note that the cylinder has a flat surface at either end, as well as a curved surface.)
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