Problem 39 - University Physics with Modern Physics

Question

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

Step-by-Step Solution

Answer

The unit vectors are: (a) \(0\hat{\boldsymbol{\imath}} - \hat{\boldsymbol{\jmath}}\), (b) \(0.7071\hat{\boldsymbol{\imath}} + 0.7071\hat{\boldsymbol{\jmath}}\), (c) \(-0.389\hat{\boldsymbol{\imath}} + 0.919\hat{\boldsymbol{\jmath}}\).

Step 1: Introduction to the Problem

We need to find the unit vector \(\mathbf{r}\) in the direction of a field point from the origin. This requires finding the vector from the origin to the given point and then determining the unit vector, which is the vector divided by its magnitude.

Step 2: Calculate the vector and its magnitude for Point a

For point (a), the coordinates are \((0, -1.35 \text{ m})\). The vector \(\mathbf{r}\) from the origin to this point is \(0\hat{\boldsymbol{\imath}} - 1.35\hat{\boldsymbol{\jmath}}\). The magnitude of this vector is given by \( |\mathbf{r}| = \sqrt{0^2 + (-1.35)^2} = 1.35 \text{ m} \).

Step 3: Find the unit vector for Point a

The unit vector \(\mathbf{r}_a\) in the direction of point (a) is \( \frac{0\hat{\boldsymbol{\imath}} - 1.35\hat{\boldsymbol{\jmath}}}{1.35} = 0\hat{\boldsymbol{\imath}} - \hat{\boldsymbol{\jmath}} \).

Step 4: Calculate the vector and its magnitude for Point b

For point (b), the coordinates are \((0.12 \text{ m}, 0.12 \text{ m})\). The vector \(\mathbf{r}\) is \(0.12\hat{\boldsymbol{\imath}} + 0.12\hat{\boldsymbol{\jmath}}\). The magnitude is \(|\mathbf{r}| = \sqrt{(0.12)^2 + (0.12)^2} = 0.1697 \text{ m} \).

Step 5: Find the unit vector for Point b

The unit vector \(\mathbf{r}_b\) is \( \frac{0.12\hat{\boldsymbol{\imath}} + 0.12\hat{\boldsymbol{\jmath}}}{0.1697} = 0.7071\hat{\boldsymbol{\imath}} + 0.7071\hat{\boldsymbol{\jmath}} \).

Step 6: Calculate the vector and its magnitude for Point c

For point (c), the coordinates are \((-1.10 \text{ m}, 2.60 \text{ m})\). The vector \(\mathbf{r}\) is \(-1.10\hat{\boldsymbol{\imath}} + 2.60\hat{\boldsymbol{\jmath}}\). The magnitude is \(|\mathbf{r}| = \sqrt{(-1.10)^2 + (2.60)^2} = 2.828 \text{ m} \).

Step 7: Find the unit vector for Point c

The unit vector \(\mathbf{r}_c\) is \( \frac{-1.10\hat{\boldsymbol{\imath}} + 2.60\hat{\boldsymbol{\jmath}}}{2.828} = -0.389\hat{\boldsymbol{\imath}} + 0.919\hat{\boldsymbol{\jmath}} \).

Key Concepts

Point Charge

A point charge is a fundamental concept in electrostatics. It represents a charge located at a single point in space. Imagine an infinitesimally small particle that holds a certain electric charge.
This concept is essential because it simplifies complex problems by allowing us to focus on the electric effects from a single source point. All calculations involving point charges use the position of the charge, which is often given as coordinates such as \(x,y\).
In practical terms, understanding point charges helps in analyzing electric fields. The field created by a point charge is radial and its strength diminishes with increased distance from the source.
This simplification becomes especially useful in learning charge interactions and electric field lines.

Vector Magnitude

The vector magnitude is an important characteristic that represents the length of a vector.
In our exercise, a vector was defined by its direction and magnitude, originating from the point charge and pointed towards a field point.
To find the magnitude, we apply the Pythagorean theorem to the vector components. For a vector \((x, y)\), the magnitude is calculated as \(|\mathbf{r}| = \sqrt{x^2 + y^2}\).
This gives us a scalar value that describes how far the field point is from the origin, or point charge, without considering direction.
Understanding and calculating vector magnitude is a key skill in physics and engineering, as it helps quantify directional quantities such as force, velocity, or in this case, the electric field.

Unit Vector Calculation

Calculating a unit vector involves transforming any vector into a vector of length 1, preserving its original direction. This is crucial in pinpointing direction in space.
The process begins by dividing each component of a vector by its magnitude. This standardizes the vector length to 1 while keeping its trajectory the same.
For our original exercise, this process was done for vectors ending at various field points. By normalizing vectors like \(0 \, \hat{\imath} - 1.35 \, \hat{\jmath}\), the result was \(0 \, \hat{\imath} - 1 \, \hat{\jmath}\).
The steps are simple:

• Calculate the vector's magnitude.
• Divide each vector component by this magnitude.

Unit vectors are vital in geometric and physical calculations because they allow for uniform scaling without altering vector directions.

Coordinate Geometry

Coordinate geometry, also known as analytic geometry, merges algebraic equations and geometric figures.
It enables us to define points, lines, and curves in the space using coordinate systems like the Cartesian plane.
The exercise leverages \(x, y\) coordinates to locate field points relative to the point charge at the origin. This is a practical application of coordinate geometry, giving a precise method to map positions and compute directions or distances between points.
Using coordinates, we easily compute the exact vector direction and magnitude that are crucial for determining unit vectors.
Coordinate geometry is foundational in many fields. It creates a bridge between mathematical theory and real-world application, helping solve spatial problems systematically and accurately.