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A point change is at the origin. With this point charge as the source point, what is the unit vector \(\hat{r}\) in the direction of \((a)\) the field point at \(x=0, y=-1.35 \mathrm{m} ;(b)\) the field point at \(x=\) \(12.0 \mathrm{cm}, y=12.0 \mathrm{cm} ;(c)\) the field point at \(x=-1.10 \mathrm{m}, y=\) 2.60 \(\mathrm{m} ?\) Express your results in terms of the unit vectors \(\hat{i}\) and \(\hat{j}\) .

Short Answer

Expert verified
(a) \(0\hat{i} - 1\hat{j}\); (b) \(0.707\hat{i} + 0.707\hat{j}\); (c) \(-0.390\hat{i} + 0.922\hat{j}\)."

Step by step solution

01

Understand the Problem

We are given a point charge at the origin and are tasked with finding the unit vector \( \hat{r} \) in the direction of the field point given specific \( x \) and \( y \) coordinates. The unit vector indicates direction and has a magnitude of 1.
02

Find the Vector Coordinates

To find the direction vector from the origin to any field point, we simply use the coordinates of the field point itself, \( \mathbf{r} = x\hat{i} + y\hat{j} \), where \( x \) and \( y \) are the coordinates of the field point.
03

Calculate Magnitude of Vector

For each coordinated point, calculate the magnitude of the vector \( \mathbf{r} \) using the formula: \[ \| \mathbf{r} \| = \sqrt{x^2 + y^2} \].
04

Determine Unit Vector

The unit vector \( \hat{r} \) is given by dividing the vector by its magnitude: \[ \hat{r} = \frac{\mathbf{r}}{\|\mathbf{r}\|} = \frac{x}{\|\mathbf{r}\|}\hat{i} + \frac{y}{\|\mathbf{r}\|}\hat{j} \].
05

Calculate for Coordinates (0, -1.35 m)

1. Vector \( \mathbf{r} = 0\hat{i} - 1.35\hat{j} \) 2. Magnitude \( \| \mathbf{r} \| = \sqrt{0^2 + (-1.35)^2} = 1.35 \) 3. Unit vector \( \hat{r} = \frac{0}{1.35}\hat{i} + \frac{-1.35}{1.35}\hat{j} = 0\hat{i} - 1\hat{j} \).
06

Calculate for Coordinates (12.0 cm, 12.0 cm)

1. Convert units: \( 12.0 \text{ cm} = 0.12 \text{ m} \) 2. Vector \( \mathbf{r} = 0.12\hat{i} + 0.12\hat{j} \) 3. Magnitude \( \| \mathbf{r} \| = \sqrt{0.12^2 + 0.12^2} = 0.17 \) 4. Unit vector \( \hat{r} = \frac{0.12}{0.17}\hat{i} + \frac{0.12}{0.17}\hat{j} = 0.707\hat{i} + 0.707\hat{j} \).
07

Calculate for Coordinates (-1.10 m, 2.60 m)

1. Vector \( \mathbf{r} = -1.10\hat{i} + 2.60\hat{j} \) 2. Magnitude \( \| \mathbf{r} \| = \sqrt{(-1.10)^2 + (2.60)^2} = 2.82 \) 3. Unit vector \( \hat{r} = \frac{-1.10}{2.82}\hat{i} + \frac{2.60}{2.82}\hat{j} = -0.390\hat{i} + 0.922\hat{j} \).

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Key Concepts

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

Vector Calculus
Vector calculus plays a crucial role in understanding electromagnetic fields, especially when dealing with problems involving vectors. It allows us to calculate the direction and magnitude of forces, fields, and points in physics. Vectors are mathematical entities that have both magnitude and direction. In the context of electromagnetic fields, vectors can represent quantities such as electric fields and magnetic fields.

When dealing with vectors, we often use components such as the x-component or the y-component, which are represented by unit vectors like \( \hat{i} \) and \( \hat{j} \). For example, if we have a vector \( \mathbf{r} = x \hat{i} + y \hat{j} \), \( x \) and \( y \) are the vector's components. These components are essential when calculating the direction of fields or forces from a point charge.

The concepts of vector addition, subtraction, and scaling are fundamental operations in vector calculus used to solve physics problems in this area. Vector calculus allows us to transform complex multidimensional problems into simpler, solvable equations. Understanding these mathematical principles helps in solving practical physics problems effectively.
Unit Vector Calculation
A unit vector is a vector with a magnitude of one. It is employed to denote direction. The unit vector in the context of electromagnetic fields shows the direction of the field lines emanating from or entering a point charge. Calculating the unit vector of a vector \( \mathbf{r} \) involves dividing each component of the vector by its magnitude.

Here's a step-by-step way to understand the calculation:
  • First, extract the vector's components based on the coordinates of a field point. For example, the field point at \( x = 0, y = -1.35 \mathrm{m} \) gives us the vector \( \mathbf{r} = 0 \hat{i} - 1.35 \hat{j} \).
  • Next, compute the magnitude of the vector using the formula \( \| \mathbf{r} \| = \sqrt{x^2 + y^2} \). This gives us a clear understanding of the vector's length or magnitude.
  • Finally, divide each component of the vector by its magnitude to find the unit vector \( \hat{r} = \frac{x}{\|\mathbf{r}\|} \hat{i} + \frac{y}{\|\mathbf{r}\|} \hat{j} \).
The resulting unit vector provides a directional component from the origin toward the field point, simplifying vector representations.
Physics Problem Solving
Physics problems involving vector calculus require a structured approach to ensure correct results. Typically, a physics problem-solving strategy in this context involves several essential steps.

Start by understanding what the problem is asking. Identify the knowns and unknowns and visualize the physical context. This helps in mapping out the direction and position of vectors.

Developing an approach involves using appropriate mathematical principles or formulas, like calculating the magnitude of vectors, then substituting values into the vector equations. Being detail-oriented during this step helps avoid errors.

Applying these strategies lets you solve various physics problems, such as determining electromagnetic field directions. Favor a systematic method that converts complex equations into simpler vector expressions, ensuring accurate results and deepening your grasp of physics concepts.
Coordinate Systems in Physics
Coordinate systems are integral to solving physics problems related to vector calculus and electromagnetic fields. They provide a reference framework for describing the orientations and positions of vectors.

The most common coordinate system in use is the Cartesian coordinate system, which is simple and intuitive. It consists of the x, y, and z axes that correspond to the directions of \( \hat{i} \), \( \hat{j} \), and \( \hat{k} \), respectively. This system allows the easy definition and manipulation of vectors in space.

To efficiently use these systems in physics problems:
  • Identify the relevant axes and align your vectors accordingly. For instance, use the \( \hat{i} \) and \( \hat{j} \) unit vectors for problems in 2D space involving point charges.
  • Map out the vectors in relation to the origin or reference point, like a point charge at the origin, to visualize direction and magnitude clearly.
  • Use coordinate transformations where necessary to switch between different systems, which aids in simplifying the problem-solving process.
Mastering these systems provides a foundation for advanced physics problem-solving involving electromagnetic fields.

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Most popular questions from this chapter

Two tiny balls of mass \(m\) carry equal but opposite charges of magnitude \(q\) . They are tied to the same ceiling hook by light strings of length L. When a horizontal uniform electric field \(E\) is turned on, the balls hang with an angle \(\theta\) between the strings (Fig, 21.46\()\) . (a) Which ball (the right or the left) is positive, and which is negative? (b) Find the angle \(\theta\) between the strings in terms of \(E, q, m,\) and \(g\) . (c) As the electric field is gradually increased in strength, what does your result from part (b) give for the largest possible angle \(\theta ?\)

A straight, nonconducting plastic wire 8.50 \(\mathrm{cm}\) long carries a charge density of \(+175 \mathrm{nC} / \mathrm{m}\) distributed uniformly along its length. It is lying on a horizontal tabletop. (a) Find the magnitude and direction of the electric field this wire produces at a point 6.00 \(\mathrm{cm}\) directly above its midpoint. (b) If the wire is now bent into a circle lying flat on the table, find the magnitude and direction of the electric field it produces at a point 6.00 \(\mathrm{cm}\) directly above its center.

Operation of an Inkjet Printer. In an inkjet printer, letters are built up by squirting drops of ink at the paper from a rapidly moving nozzle. The ink drops, which have a mass of \(1.4 \times 10^{-8} \mathrm{g}\) each, leave the nozzle and travel toward the paper at 20 \(\mathrm{m} / \mathrm{s}\) , passing through a charging unit that gives each drop a positive charge \(q\) by removing some electrons from it. The drops then pass between parallel deflecting plates 2.0 \(\mathrm{cm}\) long where there is a uniform vertical electric field with magnitude \(8.0 \times 10^{4} \mathrm{N} / \mathrm{C}\) . If a drop is to be deflected 0.30 \(\mathrm{mm}\) by the time it reaches the end of the deflection plates, what magnitude of charge must be given to the drop?

Two positive point charges \(q\) are placed on the \(y\) -axis at \(y=a\) and \(y=-a .\) A negative point charge \(-Q\) is located at some point on the \(+x\) -axis. (a) In a free-body diagram, show the forces that act on the charge \(-Q .\) (b) Find the \(x\) -and \(y\) -components of the net force that the two positive charges exert on \(-Q\) . (Your answer \(r\) . should involve only \(k, q, Q, a\) and the coordinate \(x\) of the third charge.) (c) What is the net force on the charge \(-Q\) when it is at the origin \((x=0) ?(\mathrm{d})\) Graph the \(x\) -component of the net force on the charge \(-Q\) as a function of \(x\) for values of \(x\) between \(-4 a\) and \(+4 a\) .

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.

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