Problem 17 - University Physics with Modern Physics

Question

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

Step-by-Step Solution

Answer

At (a) 0.200 m, the net electric field is 2.16 x 10^5 N/C upward. At (b) 0.600 m, the net electric field is 0.72 x 10^5 N/C upward.

Step 1: Understand the Concept

We need to find the net electric field due to two parallel lines of charge at specific points along the y-axis. Each line of charge produces an electric field, which we will calculate separately.

Step 2: Identify the Electric Field Due to a Line of Charge

The electric field created by an infinitely long line of charge at distance \(r\) is given by \(E = \frac{2k \lambda}{r}\), where \(k\) is Coulomb's constant (\(8.99 \times 10^9\, N \cdot m^2/C^2\)) and \(\lambda\) is the charge per unit length.

Step 3: Calculate the Electric Field at Point (a) y = 0.200 m

For point (a), which is at \(y = 0.200\, m\), calculate the distance from each line. For the first line at \(y = 0\), \(r = 0.200\, m\). For the second line at \(y = 0.400\, m\), \(r = 0.200\, m\). Calculate the electric fields:- First line: \(E_1 = \frac{2 \times 8.99 \times 10^9 \times 4.80 \times 10^{-6}}{0.200} = 4.32 \times 10^5\, N/C\) upward.- Second line: \(E_2 = \frac{2 \times 8.99 \times 10^9 \times (-2.40 \times 10^{-6})}{0.200} = -2.16 \times 10^5\, N/C\) downward.

Step 4: Superpose the Electric Fields for Point (a)

Add the fields vectorially. Since the fields are on the y-axis, simply add their magnitudes considering their direction:- Net field at (a): \(E_{net} = 4.32 \times 10^5 \,- 2.16 \times 10^5 = 2.16 \times 10^5\, N/C\), directed upward.

Step 5: Calculate the Electric Field at Point (b) y = 0.600 m

For point (b), which is at \(y = 0.600\, m\):- Distance from first line \(r = 0.600\, m\).- Distance from the second line \(r = 0.200\, m\).Calculate the electric fields:- First line: \(E_1 = \frac{2 \times 8.99 \times 10^9 \times 4.80 \times 10^{-6}}{0.600} = 1.44 \times 10^5\, N/C\) downward.- Second line: \(E_2 = \frac{2 \times 8.99 \times 10^9 \times (-2.40 \times 10^{-6})}{0.200} = -2.16 \times 10^5\, N/C\) upward.

Step 6: Superpose the Electric Fields for Point (b)

Add the fields vectorially:- Net field at (b): \(E_{net} = -1.44 \times 10^5 + 2.16 \times 10^5 = 0.72 \times 10^5\, N/C\), directed upward.

Key Concepts

Charge per Unit Length

Charge per unit length is a fundamental concept when dealing with continuous distributions of charge, such as lines or surfaces of charge. In simpler terms, it describes how much charge is found per unit of length along a charged object. It is often denoted by the symbol \( \lambda \) and is measured in units of Coulombs per meter (C/m).

Understanding charge per unit length helps us to handle and calculate the electric fields resulting from line charges effectively. For instance, in the original exercise, we see lines of charge with charge densities of 4.80 \( \mu \)C/m and -2.40 \( \mu \)C/m. This means that for every meter along these lines, there are 4.80 microcoulombs of charge distributed continuously along one line, and -2.40 microcoulombs along the other.

For practical purposes, this concept allows us to approximate the effect of electric fields as if the lines were composed of infinitely small charged segments. This makes it easier to apply other principles like Coulomb's Law and the superposition principle, resulting in computed electric fields that align with experimental observations.

Superposition Principle

The superposition principle is a straightforward but powerful tool in the world of physics. It states that when multiple forces or fields act on a point, the total force or field at that point is simply the vector sum of all individual forces or fields. In the context of electric fields, this means that if several electric fields interact at a point, the resultant field is just the sum of all those fields.

This principle is crucial for solving problems with multiple charges or distributions, like the one in our exercise. Each line of charge creates its own electric field at a given point, and these fields can be added together to find the net effect. The key is to first calculate the electric field each line produces individually and then add them together, taking care of the direction for each field.

• Vector Addition: Since electric fields have both magnitude and direction, they must be added as vectors. This means considering the direction each field points when computing the net field.
• Opposite Charges: Different charges can produce fields in opposite directions, either reinforcing or cancelling each other out.

This principle makes it feasible to analyze complex systems by breaking them down into simpler parts, making calculations manageable and intuitive.

Coulomb's Law

Coulomb's Law is a pivotal concept in electromagnetism describing the force between two point charges. Specifically, it quantifies the electric force as directly proportional to the product of the magnitudes of the two charges and inversely proportional to the square of the distance between them. Mathematically, it is expressed as:\[F = k \frac{|q_1 q_2|}{r^2}\]where \( F \) is the magnitude of the force, \( q_1 \) and \( q_2 \) are the point charges, \( r \) is the distance between the charges, and \( k \) is Coulomb's constant (\( 8.99 \times 10^9 \, N \cdot m^2/C^2 \)).

Though it's typically used for point charges, Coulomb's Law is integral for computing the electric field from charged lines, surfaces, and other distributions. By integrating the contributions from all small charge elements within a distribution, one can apply Coulomb's Law continuously to evaluate the field at any point.

• Application to line charges: Utilizing the concept of charge per unit length and integrating along the line leads to a formula to determine the electric field caused by an entire charged line.
• Importance of distance: As the field strength diminishes with distance, this relationship helps predict how far-reaching the influence of a charge can be.

Coulomb’s Law thus becomes a cornerstone for understanding electric fields, electric forces, and allows us to make accurate predictions in electrostatic scenarios.