Problem 21 - University Physics with Modern Physics

Question

22.21. A very long uniform line of charge has charge per unit length 4.80\(\mu \mathrm{C} / \mathrm{m}\) and lies along the \(x\) -axis. A second long uniform line of charge has charge per unit length \(-2.40 \mu \mathrm{C} / \mathrm{m}\) and is parallel to the \(x\) -axis at \(y=0.400 \mathrm{m}\) . What is the net clectric field (magnitude and direction) at the following points on the \(y\) -axis: (a) \(y=0.200 \mathrm{m}\) and \((\mathrm{b}) y=0.600 \mathrm{m} ?\)

Step-by-Step Solution

Answer

At \( y=0.200 \mathrm{m} \), the net electric field is \( E_1 - E_2 \) pointing away from the first line. At \( y=0.600 \mathrm{m} \), it is \( E_2 - E_1 \) pointing towards the second line.

Step 1: Identify the Given Parameters

We are given two parallel lines of charge. The first has a charge per unit length \( \lambda_1 = 4.80 \mu \mathrm{C/m} = 4.80 \times 10^{-6} \mathrm{C/m} \) and is located along the \( x \)-axis. The second has a charge per unit length \( \lambda_2 = -2.40 \mu \mathrm{C/m} = -2.40 \times 10^{-6} \mathrm{C/m} \) and is parallel at a distance \( 0.400 \mathrm{m} \) along the \( y \)-axis.

Step 2: Understand the Electric Field Due to a Line Charge

The electric field due to a line charge at a perpendicular distance \( r \) from the line is given by the expression \( E = \frac{\lambda}{2\pi \varepsilon_0 r} \), where \( \varepsilon_0 = 8.85 \times 10^{-12} \mathrm{C^2/N \, m^2} \) is the permittivity of free space.

Step 3: Calculate Electric Field at \( y = 0.200 \mathrm{m} \) Due to Each Line

At \( y = 0.200 \mathrm{m} \), point P is \( 0.200 \mathrm{m} \) from the first line and \( 0.400 - 0.200 = 0.200 \mathrm{m} \) from the second line. Therefore, \( E_1 = \frac{4.80 \times 10^{-6}}{2\pi \times 8.85 \times 10^{-12} \times 0.200} \) \( E_2 = \frac{-2.40 \times 10^{-6}}{2\pi \times 8.85 \times 10^{-12} \times 0.200} \). Calculate these expressions to find the magnitudes.

Step 4: Determine Net Electric Field at \( y = 0.200 \mathrm{m} \)

Since both fields point in the direction away from their respective lines of charge, calculate the net electric field by adding these components algebraically: \( E_{ ext{net, at } y=0.200} = E_1 - E_2 \) as they are in opposite directions.

Step 5: Calculate Electric Field at \( y = 0.600 \mathrm{m} \) Due to Each Line

At \( y = 0.600 \mathrm{m} \), point Q is \( 0.600 \mathrm{m} \) from the first line and \( 0.600 - 0.400 = 0.200 \mathrm{m} \) from the second line. Thus, \( E_1 = \frac{4.80 \times 10^{-6}}{2\pi \times 8.85 \times 10^{-12} \times 0.600} \) \( E_2 = \frac{-2.40 \times 10^{-6}}{2\pi \times 8.85 \times 10^{-12} \times 0.200} \). Calculate these expressions to find the magnitudes.

Step 6: Determine Net Electric Field at \( y = 0.600 \mathrm{m} \)

Using the calculated magnitudes, determine the net electric field at this point using vector sum: \( E_{ ext{net, at } y=0.600} = E_2 - E_1 \) because the direction of the fields due to \( \lambda_2 \) becomes dominant at this greater distance.

Key Concepts

Line Charge

A line charge refers to a charge distributed uniformly along a line. This type of charge distribution is common in theoretical physics because it allows us to simplify the calculation of the electric field due to the charge configuration. A line charge is described by its linear charge density, denoted as \( \lambda \), which is the charge per unit length of the line. In this case, we have two line charges: one with \( \lambda_1 = 4.80 \, \mu \mathrm{C/m} \) and another with \( \lambda_2 = -2.40 \, \mu \mathrm{C/m} \).
These values indicate two lines with different charge intensities:

• Positive line charge: This will create an electric field pointing away from the line.
• Negative line charge: This will produce an electric field towards the line, owing to its negative nature.

The concept here is critical because the behavior of these electric fields affects points in the surrounding space such as the \( y \)-axis points mentioned in the exercise.

Permittivity of Free Space

Permittivity of free space, represented by \( \varepsilon_0 \), is a fundamental constant in electromagnetism. It is used to quantify the ability of the vacuum to permit electric field lines. The value of \( \varepsilon_0 \) is approximately \( 8.85 \times 10^{-12} \, \mathrm{C^2/N \, m^2} \). This constant appears in formulas that describe how electric fields interact with materials and charges.
In the context of a line charge, the electric field \( E \) at a distance \( r \) from a line charge of linear charge density \( \lambda \) can be expressed as:\[ E = \frac{\lambda}{2\pi \varepsilon_0 r} \]This formula shows that the electric field intensity decreases with distance from the line charge, but also depends linearly on the charge density and inversely on the permittivity of free space.
Understanding the role of \( \varepsilon_0 \) helps grasp how electric fields propagate through space and interact with different charge distributions. It provides the medium's response to the presence of an electric field, thus allowing electric line charges to impact different points in surrounding space.

Superposition of Electric Fields

The superposition principle is a foundational concept in physics that allows the analysis of multiple influences on a given system by simply adding the effects produced by each influence separately. When dealing with electric fields, you can calculate the total electric field created by multiple charges by summing the electric fields produced by each charge individually. This is particularly useful when assessing the net electric field at specific points due to complex charge configurations.
In the exercise, we have two line charges with opposite signs:

• Line charge \( \lambda_1 \) creates an electric field \( E_1 \).
• Line charge \( \lambda_2 \) creates an electric field \( E_2 \).

To find the net electrical field at any point on the \( y \)-axis, like \( y=0.200 \, \mathrm{m} \) or \( y=0.600 \, \mathrm{m} \), we simply add the individual electric fields considering their directions. The result is a vector sum. At point \( y=0.200 \, \mathrm{m} \), for instance, both fields contribute in directions relative to their lines:

• If both fields point away from or towards their origin, they are summed.
• Otherwise, their magnitudes are subtracted if they are oppositely directed.

The principle of superposition is immensely practical as it simplifies complex systems into more manageable parts, allowing for easier computation and understanding.