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?

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: $$ E_1 = E_2 \approx 1.19 \times 10^{5} \mathrm{N/C} $$

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: $$ E_{net} = \sqrt{E_{netx}^2 + E_{nety}^2} $$ $$ E_{net} \approx 1.18 \times 10^{5} \mathrm{N/C} $$

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.