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

Short Answer

Expert verified
(a) \(\hat{r} = -\hat{\jmath}\); (b) \(\hat{r} = 0.707\hat{\imath} + 0.707\hat{\jmath}\); (c) \(\hat{r} \approx -0.389\hat{\imath} + 0.920\hat{\jmath}\)."

Step by step solution

01

Vector from Source to Field Point (a)

The vector from the origin to the point \((0, -1.35)\) is simply \(\mathbf{r} = 0\hat{\imath} - 1.35\hat{\jmath}\).
02

Calculate Magnitude of Vector (a)

The magnitude of the vector \(\mathbf{r}\) is given by \(|\mathbf{r}| = \sqrt{0^2 + (-1.35)^2} = 1.35\).
03

Determine Unit Vector (a)

The unit vector \(\hat{\mathbf{r}}\) is found by dividing the original vector by its magnitude: \(\hat{\mathbf{r}} = \frac{0\hat{\imath} - 1.35\hat{\jmath}}{1.35} = 0\hat{\imath} - \hat{\jmath}\).
04

Vector from Source to Field Point (b)

The vector from the origin to the point \((0.12, 0.12)\) is \(\mathbf{r} = 0.12\hat{\imath} + 0.12\hat{\jmath}\).
05

Calculate Magnitude of Vector (b)

The magnitude of the vector \(\mathbf{r}\) is \(|\mathbf{r}| = \sqrt{(0.12)^2 + (0.12)^2} = \sqrt{0.0288} = 0.1697\).
06

Determine Unit Vector (b)

The unit vector \(\hat{\mathbf{r}}\) is \(\hat{\mathbf{r}} = \frac{0.12\hat{\imath} + 0.12\hat{\jmath}}{0.1697} = 0.707\hat{\imath} + 0.707\hat{\jmath}\).
07

Vector from Source to Field Point (c)

The vector from the origin to the point \((-1.10, 2.60)\) is \(\mathbf{r} = -1.10\hat{\imath} + 2.60\hat{\jmath}\).
08

Calculate Magnitude of Vector (c)

The magnitude is \(|\mathbf{r}| = \sqrt{(-1.10)^2 + (2.60)^2} = \sqrt{1.21 + 6.76} = \sqrt{7.97} = 2.824\).
09

Determine Unit Vector (c)

The unit vector \(\hat{\mathbf{r}}\) is \(\hat{\mathbf{r}} = \frac{-1.10\hat{\imath} + 2.60\hat{\jmath}}{2.824} \approx -0.389\hat{\imath} + 0.920\hat{\jmath}\).

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Key Concepts

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

Point Charge
In electrostatics, a point charge refers to a model or an idealization of a charged particle where the charge is assumed to be concentrated at a single point in space. This model is pivotal for simplifying the analysis and calculation of electric fields, specifically when the charges are considered very small compared to the distances involved.

Imagine a tiny sphere that holds an electric charge; we disregard its size for simplicity by treating it as a point. This approximation helps in deriving various electric field formulas since it's easier to calculate the effects coming from a single point in space.

Such point charges are often used as source points in electrostatics, where the origin of a coordinate system is usually chosen for ease of computation. You can think about it like starting point at zero from which you measure other positions.
Unit Vector
A unit vector is a concept used in vector mathematics where a vector has a length, or magnitude, of one unit. It serves as a direction marker in vector spaces, indicating the direction of a vector without considering its magnitude.

When you derive a unit vector from another vector, you're essentially scaling the vector down to a magnitude of 1 while preserving its direction. For example, if you have a vector \(\mathbf{r}\), to find the unit vector \(\hat{\mathbf{r}}\), you divide each component of \(\mathbf{r}\) by its magnitude:
\[ \hat{\mathbf{r}} = \frac{\mathbf{r}}{\|\mathbf{r}\|}\]

This unit vector is frequently used in physics and engineering to specify directions. It helps, in particular, to identify the direction of electric field vectors originating due to point charges.
Vector Magnitude
The magnitude of a vector is a measure of its length irrespective of its direction. Essentially, it captures how "long" the vector is or how far it "reaches" in space.

Given a vector \(\mathbf{r} = a\hat{\imath} + b\hat{\jmath}\), its magnitude is calculated using the Pythagorean theorem:
\[ \|\mathbf{r}\| = \sqrt{a^2 + b^2}\]
This formula sums the squares of its components, reflecting the vector's distance from origin in terms of its length. The magnitude is crucial for converting vectors into unit vectors by acting as the divisor so each component of the vector is scaled to 1 unit in length.
Coordinate System
A coordinate system is an agreement or framework to quantify the position of points in space. In physics, especially in studies like electrostatics, it aids in determining scalar and vector quantities' positions.

The Cartesian coordinate system is among the most straightforward and widely-used systems. It uses perpendicular axes—typically denoted as \(x\) and \(y\) in two-dimensional space, \(z\) is added for three-dimensional space—to describe any point's location by its distance from these perpendicular lines.
  • The origin is the point where all axis values are zero (0, 0 for two dimensions).
  • Vectors are often described in terms of their beginning and endpoint within a coordinate system.
By using this system, the location and direction of point charges and vectors become clearer, facilitating easy computation of vector-related tasks, such as determining unit vectors or magnitudes.

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Most popular questions from this chapter

You have a pure (24-karat) gold ring of mass 10.8 g. Gold has an atomic mass of 197 g\(/\)mol and an atomic number of 79. (a) How many protons are in the ring, and what is their total positive charge? (b) If the ring carries no net charge, how many electrons are in it?

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