Problem 50 - University Physics with Modern Physics

Question

A point charge \(q_{1}=-4.00 \mathrm{nC}\) is at the point \(x=\) \(0.600 \mathrm{m}, y=0.800 \mathrm{m},\) and a second point charge \(q_{2}=+6.00 \mathrm{nC}\) is at the point \(x=0.600 \mathrm{m}, y=0 .\) Calculate the magnitude and direction of the net electric field at the origin due to these two point charges.

Step-by-Step Solution

Answer

The net electric field at the origin is approximately \( 98.9 \text{ N/C} \) at an angle of \(-77.5^\circ\) from the positive x-axis.

Step 1: Identify Positions of Charges and the Point of Interest

Charge \( q_1 = -4.00 \text{ nC} \) is located at \((x_1, y_1) = (0.600, 0.800) \text{ m}\), and charge \( q_2 = +6.00 \text{ nC} \) is located at \((x_2, y_2) = (0.600, 0.0) \text{ m}\). The electric field is to be calculated at the origin \((0, 0)\).

Step 2: Calculate the Distance to Each Charge

Calculate the distance from the origin to each charge. For \( q_1 \), the distance is \( r_1 = \sqrt{(0.600)^2 + (0.800)^2} \). For \( q_2 \), the distance is \( r_2 = \sqrt{(0.600)^2 + (0)^2} = 0.600 \text{ m}\).

Step 3: Calculate the Electric Field Due to Each Charge

The magnitude of the electric field due to a point charge is given by \( E = \frac{k |q|}{r^2} \), where \( k = 8.99 \times 10^9 \text{ N m}^2/\text{C}^2 \). Calculate \( E_1 \) due to \( q_1 \) and \( E_2 \) due to \( q_2 \).

Step 4: Determine the Components of Each Electric Field

Since \( q_1 \) is at \( (0.600, 0.800) \), its electric field, \( E_1 \), can be broken into components: \( E_{1x} = E_1 \cos(\theta_1) \) and \( E_{1y} = E_1 \sin(\theta_1) \), where \( \theta_1 = \tan^{-1}(\frac{0.800}{0.600}) \). Similarly, \( E_{2y} = 0 \) because \( q_2 \) is along the y-axis, and \( E_{2x} = E_2 \) because it is along the x-axis.

Step 5: Calculate the Net Electric Field Components

Find the net components: \( E_{net,x} = E_{1x} + E_{2x} \) and \( E_{net,y} = E_{1y} + E_{2y} \). Consider the signs of charges when calculating the direction of electric fields.

Step 6: Calculate the Magnitude and Direction of the Net Electric Field

The magnitude of the net electric field is given by \( E_{net} = \sqrt{E_{net,x}^2 + E_{net,y}^2} \). The direction is \( \theta = \tan^{-1}\left(\frac{E_{net,y}}{E_{net,x}}\right) \).

Step 7: Solve for Specific Values

Use the given charges and positions to compute specific values: \( r_1 = 1 \text{ m} \), \( E_1 = 8.99 \times 10^9 \times \frac{4.00 \times 10^{-9}}{(1)^2} \), and \( E_2 = 8.99 \times 10^9 \times \frac{6.00 \times 10^{-9}}{(0.600)^2} \). Calculate components and find \( E_{net} \) and direction \( \theta \).

Step 8: Final Calculation and Result

Compute the numerical values for the electric field components and combine them to find the exact magnitude and direction at the origin. Use the results from the previous steps to summarize the net field at the origin.

Key Concepts

Point Charges

Point charges are small charged particles considered to be at a single point in space. They serve as useful idealized models in physics to study electric fields and forces. In the given exercise, we are dealing with two point charges, \( q_1 \) and \( q_2 \), each with a specific charge and location. These charges create an electric field around them, due to the electric forces they exert on other charges nearby. Understanding the position of these charges is essential because the electric field strength depends on the distance between the point of interest and the charges.
For example, \( q_1 = -4.00 \text{ nC} \) is located at \((0.600, 0.800)\) meters, while \( q_2 = +6.00 \text{ nC} \) is located at \((0.600, 0.0)\) meters. Calculating the electric field involves determining the individual contribution from each charge at a specific point, like the origin.

Electric Field Components

The electric field produced by a point charge at a certain distance can be quantified as a vector with both magnitude and direction. To effectively analyze the net field from multiple charges, it's vital to break down their fields into components. Each component aligns with either the x-axis or y-axis.
If we look at \( q_1 \), situated at \((0.600, 0.800)\), its electric field vector can be split into horizontal (x-component) and vertical (y-component) parts. The x-component, \( E_{1x} \), and the y-component, \( E_{1y} \), can be calculated using trigonometric functions as follows:

• \( E_{1x} = E_1 \cos(\theta_1) \)
• \( E_{1y} = E_1 \sin(\theta_1) \)

\( \theta_1 \) is the angle formed with the horizontal, which is determined by the position of the charge. In this case, \( \theta_1 = \tan^{-1}\left(\frac{0.800}{0.600}\right) \).
For \( q_2 \), which lies on the x-axis at \((0.600, 0)\), the electric field has no y-component, so \( E_{2y} = 0 \), making the x-component \( E_{2x} \) the same as the whole electric field \( E_2 \). These components are crucial for calculating the net electric field.

Electric Field Direction

To determine the direction of the net electric field from multiple charges, we combine the directional contributions from each charge. This involves analyzing the angles at which the electric fields act.
Using the individual components found for \( q_1 \) and \( q_2 \), we compute the net electric field components \( E_{net,x} \) and \( E_{net,y} \):

• \( E_{net,x} = E_{1x} + E_{2x} \)
• \( E_{net,y} = E_{1y} + E_{2y} \)

The direction of the net electric field is the angle \( \theta \) relative to the x-axis, given by:
\[ \theta = \tan^{-1}\left(\frac{E_{net,y}}{E_{net,x}}\right) \]
Knowing \( \theta \) tells us where the resultant field is directed in the plane. Keep in mind, the sign of the charge affects the direction of the field lines. Positive charges push away field lines, while negative charges pull them closer.

Coulomb's Law

Coulomb's law is a fundamental principle that describes the electric force between two point charges. The law states that the electric force (F) between two charges is directly proportional to the product of their magnitudes and inversely proportional to the square of the distance between them. Mathematically, it is expressed as:
\[ F = \frac{k |q_1 q_2|}{r^2} \]
where:

• \( F \) is the magnitude of the force.
• \( k = 8.99 \times 10^9 \text{ Nm}^2/\text{C}^2 \) is Coulomb's constant.
• \( |q_1| \) and \( |q_2| \) are the absolute values of the charges.
• \( r \) is the distance between the charges.

This law provides the groundwork for calculating the electric field, since the field is essentially the force per unit charge. In the exercise, Coulomb's law helps us determine the electric fields created by \( q_1 \) and \( q_2 \) at the origin. Applying the formula for each point charge, we find their respective electric fields, which are then used to compute the total field experienced at the given point.