Problem 49 - University Physics with Modern Physics

Question

In a rectangular coordinate system a positive point charge \(q=6.00 \times 10^{-9} \mathrm{Cis~placed}\) at the point \(x=+0.150 \mathrm{m}, y=0\) and an identical point charge is placed at \(x=-0.150 \mathrm{m}, y=0\) Find the \(x\) - and \(y\) -components, the magnitude, and the direction of the electric field at the following points: (a) the origin; (b) \(x=0.300 \mathrm{m}, y=0 ;(\mathrm{c}) x=0.150 \mathrm{m}, y=-0.400 \mathrm{m} ;(\mathrm{d}) x=0\) \(y=0.200 \mathrm{m}\)

Step-by-Step Solution

Answer

Calculate and sum electric field vectors for each point, considering symmetry and applying vector addition principles.

Step 1: Identify the Configuration at the Origin

For part (a), both charges are symmetrically placed on the x-axis at equal distances from the origin. Due to symmetry, the y-components of the electric fields due to each charge will cancel each other out, and only the x-components will add up. The electric field due to a single charge is given by the formula:\[ E = \frac{k \cdot |q|}{r^2} \] where \(k\) is the Coulomb's constant \( (8.99 \times 10^9 \mathrm{Nm}^2/\mathrm{C}^2) \), and \(r\) is the distance from the charge to the point in question (0.150 m). The net field at the origin is twice the x-component from one charge because both fields add along the x-axis.Using trigonometry, the x-component of the electric field due to one charge at the origin is:\[ E_x = E \cdot \cos(\theta) \]The angle \( \theta \) is 180° for both charges since they point towards the center. Thus, \(\cos(\theta) = 1 \) and, simplifying gives:\[ E_x = \frac{2k \cdot |q|}{r^2} \]

Step 2: Calculate at Point (b)

For part (b), at \( x = 0.300 \mathrm{m}, y = 0 \), both charges are again symmetrically placed along the x-axis relative to this point. The calculation follows the same steps as part (a), where we determine the contribution from each charge. Note that this point is further away from the negative x-directed charge and closer to the positive x-directed charge. We calculate the net electric field by subtracting the contributions since they are in opposite directions.For each charge:\[ E = \frac{k \cdot |q|}{(0.300 - 0.150)^2} \, + \, \frac{k \cdot |q|}{(0.300 + 0.150)^2} \]The net electric field is the absolute difference, given the opposing directions of the fields due to each charge.

Step 3: Consider Point (c)

For part (c), at the point \( x=0.150 \, \text{m},\, y=-0.400 \, \text{m} \), observe that the field contributions from each charge need to be resolved into x and y components. Calculate the distance from each charge to the point using the Pythagorean theorem:\[ r = \sqrt{(0.150 - 0.150)^2 + (0.150 + 0.400)^2} = 0.400 \, \text{m} \text{ and } \sqrt{(0.150 + 0.150)^2 + (0.150 + 0.400)^2} \]Find the electrical field for each charge separately and sum the corresponding components to find total \(E_x\) and \(E_y\). The direction is given by:\[ \theta = \arctan\left( \frac{E_y}{E_x} \right) \]

Step 4: Analyze the Field at Point (d)

For part (d), consider the point at \( x=0, y=0.200 \, \text{m} \). Calculate the contribution from each charge considering their position relative to this point. Calculate the distance using the Pythagorean theorem:\[ r = \sqrt{(0.200)^2 + (0.150)^2} \]Determine the components of the field due to each charge, sum them, and find net \(E_x\) and \(E_y\):\[ E_x = E_1 \cos(\theta) + E_2 \cos(\theta) \]\[ E_y = E_1 \sin(\theta) - E_2 \sin(\theta) \]The direction is given by:\[ \theta = \arctan\left( \frac{E_y}{E_x} \right) \]

Step 5: Calculate Magnitude and Direction

For each point computed (a to d), calculate the magnitude of the total electric field as:\[ E = \sqrt{E_x^2 + E_y^2} \]The direction (angle with respect to the x-axis) is determined as:\[ \theta = \arctan\left( \frac{E_y}{E_x} \right) \]

Key Concepts

Point Charge

A point charge refers to a charged object that is treated as if all its electric charge is concentrated at a single point in space. This simplification is useful in physics when dealing with objects where the charge distribution is much smaller than the distances involved in the problem. It allows us to apply mathematical models accurately to predict electric fields and forces.
In our given exercise, each point charge is located at specified coordinates on the x-axis. These can be imagined as tiny charged spheres positioned at exact points along this axis. The presence of these charges influences the electric field around them.
Understanding point charges helps in analyzing how they affect their surroundings, especially within a coordinate system, when calculating electric fields or forces.

Coordinate System

In the given problem, a rectangular coordinate system is used. This is a mathematical system to describe the position of a point in space using coordinates \(x, y\). It acts as a map, allowing us to locate where our point charges are situated and calculate distances between points.
For instance, the charges in the example are placed symmetrically on either side of the origin at \(x=+0.150 \, \text{m}\, \text{and} \, x=-0.150 \, \text{m}\). This symmetry makes it easier to calculate electric fields at certain points since symmetrical arrangements often lead to cancellations in some components (like the y-component in part (a) of the exercise).
The coordinate system simplifies the complex nature of electric fields generated by multiple charges, enabling us to break down the problem into comprehensible parts.

Electric Field Components

Electric fields produced by charges have both magnitude and direction. When investigating electric fields at a specific point, it's useful to resolve the field into components, typically along the x and y axes of the coordinate system.
In the exercise, for each location where the field is calculated, the vector components—each representing the influence of the field along the x or y axis—are determined. For example, at the origin (part a), the y-components cancel out due to symmetry, leaving only x-components summed.
This decomposition is crucial because it allows one to calculate the total effect of several vector fields by simply adding their components along each axis. Once the individual components are known, they can be combined to find the resultant electric field and its direction.

Coulomb's Law

Coulomb's law describes how the electric field or force between two charges is calculated. The law states that the force \(F\) or field \(E\) between two point charges is directly proportional to the product of their charges and inversely proportional to the square of the distance (r) between them:

• The formula is expressed as \( E = \frac{k \cdot |q|}{r^2} \), where \(k\) is Coulomb’s constant.
• This law explains why the magnitude of the electric field diminishes with increasing distance from a charge.

In our exercise, each step utilizes Coulomb's law to calculate the electric fields due to each point charge. The law plays a central role in understanding the behavior of electricity and magnetism by quantifying the interactions between charged particles.
By applying Coulomb's law, you can predict how a positive point charge will interact with another charge, influencing the field at a given point in space.