Question

Two parallel, uniformly charged, infinitely long wires carry opposite charges with a linear charge density \(\lambda=1.00 \mu \mathrm{C} / \mathrm{m}\) and are \(6.00 \mathrm{~cm}\) apart. What is the magnitude and direction of the electric field at a point midway between them and \(40.0 \mathrm{~cm}\) above the plane containing the two wires?

Short answer

The magnitude of the net electric field at the given point is approximately \(1.18 \times 10^{5} \mathrm{N/C}\). The direction of the electric field is along the x-axis, perpendicular to the line connecting the two wires, and away from the positive wire.

Step-by-step solution

Determine the electric field due to each wire

To begin, we need to find the electric field generated by each wire individually. For a uniformly charged, infinitely long wire, the electric field at a perpendicular distance, r, can be calculated using the formula: $$ E = \frac{1}{4\pi\epsilon_0} \cdot \frac{2 \cdot \lambda}{r} $$ where \(\epsilon_0\) is the vacuum permittivity (\(8.85 \times 10^{-12} \mathrm{C^2/Nm^2}\)), λ is the linear charge density of the wire, and r is the perpendicular distance from the wire. In our case, \(\lambda = 1.00 \times 10^{-6} \mathrm{C/m}\). The distance between the two wires is 6 cm, which means that the perpendicular distance from the midpoint is \(\frac{6}{2} = 3\) cm or 0.03 m. Substituting this value into the equation for both wires, we have: $$ E_1 = E_2 = \frac{1}{4\pi\epsilon_0} \cdot \frac{2 \cdot 1.00 \times 10^{-6}}{0.03} $$ Calculate the electric field: $$ E_1 = E_2 \approx 1.19 \times 10^{5} \mathrm{N/C} $$

Determine the net electric field at the point

To find the net electric field at the given point, we need to determine the components of the electric field due to each wire along the x and y axis. Since the wires are carrying opposite charges, the electric fields will have opposite directions along the y-axis but the same direction along the x-axis. Next, we need to calculate the x and y components of each electric field. We are given that the point is 40 cm above the plane containing the two wires or 0.4 m. So, the angle \(\theta\) between the electric field and the x-axis can be calculated using the tangent function: $$ \tan\theta = \frac{0.03}{0.4} $$ Calculate the angle: $$ \theta \approx 4.29^{\circ} $$ Now, we can find the x and y components of the electric field due to each wire (keep in mind that the y components are opposite in direction): $$ E_{1x} = E_1 \cos\theta $$ $$ E_{1y} = E_1 \sin\theta $$ Calculate the components: $$ E_{1x} = E_{2x} \approx 1.19 \times 10^{5} \cos(4.29^{\circ}) \approx 1.18 \times 10^{5} \mathrm{N/C} $$ $$ E_{1y} \approx 1.19 \times 10^{5} \sin(4.29^{\circ}) \approx 8.93 \times 10^{3} \mathrm{N/C} $$ Now, we can sum the x and y components: $$ E_{netx} = E_{1x} + E_{2x} $$ $$ E_{nety} = E_{1y} - E_{2y} =0 \mathrm{N/C} $$ Finally, calculate the magnitude of the net electric field: $$ E_{net} = \sqrt{E_{netx}^2 + E_{nety}^2} $$ $$ E_{net} \approx 1.18 \times 10^{5} \mathrm{N/C} $$

Find the direction of the net electric field

Since the net electric field has components only along the x-axis (since the y component is zero), the direction is along the horizontal plane (x-axis) and the electric field is pointing perpendicular to the line connecting the two wires and in the direction away from the positive wire. The net electric field has a magnitude of approximately \(1.18 \times 10^{5} \mathrm{N/C}\) and is directed along the x-axis, perpendicular to the line connecting the two wires and away from the positive wire.

Key concepts

Linear Charge Density

Linear charge density is a crucial concept in understanding the distribution of charge along a conductor or wire. It is defined as the amount of electric charge per unit length of the wire. In mathematical terms, it is represented by the symbol \( \lambda \) and is typically measured in coulombs per meter (C/m).

Understanding linear charge density is vital because it directly affects the strength and distribution of the electric field around the wire. The higher the charge density, the stronger the electric field at a given point away from the wire, provided the distance is the same. For example, if you have two wires with different linear charge densities, the wire with the higher density will generate a stronger electric field at the same distance. It's important to remember, in the context of our exercise, that the linear charge density for both wires is given as \(1.00 \mu \mathrm{C} / \mathrm{m}\), and this uniformity simplifies the calculation of the electric field around them.

Vacuum Permittivity

Vacuum permittivity, symbolized by \( \epsilon_0 \), is a fundamental physical constant that describes the ability of a vacuum to permit electric field lines. Also known as the electric constant, it has a value of approximately \(8.85 \times 10^{-12} \mathrm{C}^2/\mathrm{Nm}^2\) and plays an essential role in the equations for determining electric field strength around charged objects.

The constant is part of Coulomb's law, which governs the force between two point charges, and it also appears in the equation for calculating the electric field of a long charged wire. It reflects the ease with which electric field lines can permeate space or a medium – a higher permittivity allows for a denser electric field. In the exercise provided, this constant is used to calculate the electric field produced by each wire, and it's a factor that helps us understand how the vacuum itself impacts electrical interactions.

Electric Field Components

The electric field can be thought of as a vector field; it has both magnitude and direction. When dealing with complex configurations, it's often necessary to break down the electric field into components along the x, y, and z axes. This approach makes analyzing and understanding the overall field much easier.

In the situation with our two charged wires, the electric field components are calculated using trigonometric functions based on the geometry of the setup. With the wires located in a horizontal plane, the point of interest lies above this plane, so we need to calculate both the horizontal (x-axis) and vertical (y-axis) components of the electric field. In the case where the fields from two charged sources interact, the superposition principle allows us to add these vector components to calculate the net electric field at a given point. It's a technique essential for unraveling the combined effect of multiple electric fields and finding the resultant field, which in our exercise, turns out to have no vertical component and lies entirely in the horizontal direction, simplifying our visualization and understanding of this electrical phenomenon.