Problem 34 - University Physics with Modern Physics

Question

Point charge \(q_{1}=-5.00 \mathrm{nC}\) is at the origin and point charge \(q_{2}=+3.00 \mathrm{nC}\) is on the \(x\) -axis at \(x=3.00 \mathrm{cm} .\) Point \(P\) is on the \(y\) -axis at \(y=4.00 \mathrm{cm}\) . (a) Calculate the electric fields \(\vec{E}_{1}\) and \(\overrightarrow{\boldsymbol{E}}_{2}\) at point \(P\) due to the charges \(q_{1}\) and \(q_{2}\) . Express your results in terms of unit vectors (see Example \(21.6 ) .\) (b) Use the results of part (a) to obtain the resultant field at \(P\) , expressed in unit vector form.

Step-by-Step Solution

Answer

Calculate individual electric fields due to both charges and combine them to find the resultant.

Step 1: Understand the Problem

We have two point charges, \( q_1 = -5.00 \text{ nC} \) at the origin and \( q_2 = +3.00 \text{ nC} \) at \( (3.00 \text{ cm}, 0) \). We need to find the electric fields \( \vec{E}_1 \) and \( \vec{E}_2 \) at point \( P \) located at \( (0, 4.00 \text{ cm}) \). Then, we have to determine the resultant electric field at point \( P \).

Step 2: Calculate Distance and Electric Field Due to \( q_1 \)

For the charge \( q_1 \) at the origin, the distance to point \( P \) is \( 4.00 \text{ cm} \) or \( 0.04 \text{ m} \). The electric field at \( P \) is given by: \[ \vec{E}_1 = \frac{k \cdot q_1}{r^2} \hat{r} \] where \( k = 8.99 \times 10^9 \text{ N m}^2/\text{C}^2 \), and \( \hat{r} \) is the unit vector in the direction from \( q_1 \) to \( P \). Since \( q_1 \) is at the origin, the unit vector is \( \hat{j} \) (positive y-direction). Substitute the values:\[ \vec{E}_1 = \frac{8.99 \times 10^9 \times (-5.00 \times 10^{-9})}{(0.04)^2} \hat{j}\] Calculate \( \vec{E}_1 \).

Step 3: Calculate Distance and Electric Field Due to \( q_2 \)

The position of \( q_2 \) is \( (3.00 \text{ cm}, 0) \), and \( P \) is at \( (0, 4.00 \text{ cm}) \). Calculate the distance \( r \) using the Pythagorean theorem: \[ r = \sqrt{(3.00 \times 10^{-2})^2 + (4.00 \times 10^{-2})^2} = 5.00 \times 10^{-2} \text{ m} \].The electric field \( \vec{E}_2 \) at \( P \) due to \( q_2 \) is:\[ \vec{E}_2 = \frac{k \cdot q_2}{r^2} \hat{r} \] where \( \hat{r} \) is a unit vector from \( q_2 \) to \( P \). The components of \( \hat{r} \) are given by: \( \frac{-3.00 \times 10^{-2}}{5.00 \times 10^{-2}} \hat{i} + \frac{4.00 \times 10^{-2}}{5.00 \times 10^{-2}} \hat{j} \).Substitute and calculate \( \vec{E}_2 \).

Step 4: Calculate Resultant Electric Field at Point \( P \)

The resultant field \( \vec{E} \) at \( P \) is the vector sum of \( \vec{E}_1 \) and \( \vec{E}_2 \).We now add the x-components and y-components of \( \vec{E}_1 \) and \( \vec{E}_2 \). Since \( \vec{E}_1 \) has no x-component, the final components are:- \( x\)-component: \( E_{2x} \)- \( y\)-component: \( E_{1y} + E_{2y} \).Combine these components to get \( \vec{E} \).

Step 5: Express Result in Unit Vector Form

Combine the calculated x and y components to express the resultant electric field in unit vector form:\[ \vec{E} = E_{2x} \hat{i} + (E_{1y} + E_{2y}) \hat{j} \].Substitute the numerical values of \( E_{2x} \), \( E_{1y} \), and \( E_{2y} \) calculated in previous steps to obtain the final expression for \( \vec{E} \).

Key Concepts

Understanding Point Charges

Point charges are fundamental concepts in the study of electric fields. A point charge is an electric charge considered as sitting at a single point in space. These charges create electric fields, which can exert forces on other charges within the field.
In the given exercise, we have two point charges: one negative charge, \( q_1 = -5.00 \mathrm{nC} \), located at the origin, and a positive charge, \( q_2 = +3.00 \mathrm{nC} \), positioned along the \( x \)-axis. Although these charges have no physical size, they have a measureable effect on their surroundings, which can be calculated by analyzing the electric field they produce.
When analyzing point charges, it is crucial to consider both the magnitude of the charge and their location. This helps determine the direction and strength of the electric field they generate. For example, a positive charge emits field lines outwards, while a negative charge attracts these lines towards itself.

Resultant Field of Electric Forces

The resultant electric field at a point is the vector sum of all the electric fields acting at that point due to individual point charges. In this problem, we calculated the electric fields due to each charge separately at point \( P \), then combined them to find the total effect.
This concept is critical because in real-world scenarios, multiple charges affect each other simultaneously, and understanding the overall effect requires vector addition.
To find the resultant field, you need the components of each field in the \( x \)- and \( y \)-directions. Use the superposition principle to add these components:

• Add the \( x \)-components of the fields together to get the resultant \( x \)-component.
• Likewise, add the \( y \)-components to get the resultant \( y \)-component.

Combining these will give you the resultant electric field in vector form, which is crucial for understanding how multiple electric fields interact and affect a particular point in space.

Using Unit Vectors in Electric Fields

Unit vectors are dimensionless vectors with a magnitude of one used to indicate direction. They are essential in expressing vectors, like electric fields, in a standardized form.
In this exercise, the electric fields at point \( P \) due to \( q_1 \) and \( q_2 \) were expressed using unit vectors \( \hat{i} \) for the \( x \)-axis and \( \hat{j} \) for the \( y \)-axis.
The electric field due to a point charge is given in the form \( \vec{E} = E_x \hat{i} + E_y \hat{j} \). Using unit vectors helps break down the electric field into its horizontal and vertical components, making it easier to calculate and visualize.
Why are unit vectors so helpful? They simplify calculations by allowing you to easily add vector components and provide a clear way to present the direction and magnitude of vectors.