Question

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

Short answer

Answer: The magnitude of the electric field at the point midway between the two wires and 40 cm above the plane containing them is \(4.30 × 10^6 N/C\), and its direction is parallel or along the plane containing the two wires.

Step-by-step solution

Find the electric field due to each wire

The electric field due to an infinitely long, uniformly charged wire at a perpendicular distance r from the wire is given by the formula: \( E = \frac{1}{4πε_0} · \frac{2λ}{r} \) where \(ε_0\) is the vacuum permittivity, and λ is the linear charge density of the wire. In this problem, λ = \(1.00 × 10^{-6} C/m\) and \(ε_0 = 8.85 × 10^{-12} C^2/N/m^2\). The distance between the two wires is 6 cm, so the perpendicular distance from the point to each wire is 3 cm or 0.03 m. Substituting these values, we get the electric field due to each wire: \( E = \frac{1}{4π(8.85 × 10^{-12}) } · \frac{2(1.00 × 10^{-6})}{0.03} = 2.39 × 10^7 N/C \)

Find the angle between the electric fields at the specified point

Since the point is in the same perpendicular plane as the two wires, we can use the triangle formed by the point, the Midpoint of the two wires, and one of the wires to determine the angle between the electric fields. Knowing the triangle's sides: the perpendicular distance from the point to the Midpoint is 40 cm or 0.4 m, and half of the distance between the two wires is 3 cm or 0.03 m, we can calculate the angle using: \( \theta = \tan^{-1}(\frac{0.03}{0.4}) = 4.29° \)

Finding the net electric field at the point

Since both wires have the same linear charge density, the magnitude of the electric field due to each wire at the midpoint is the same. To find the net electric field at the point, we can use vector addition. Its x-components will be equal in magnitude, cancel each other out, and the net electric field will be in the y-direction (plane containing the wires). The y-component of the total electric field is the sum of the y-components of the electric fields due to each wire: \( E_y = 2(E \sin(\theta)) = 2(2.39 × 10^7 \sin(4.29°)) = 4.30 × 10^6 N/C \)

Compute the magnitude and direction of the net electric field

Since the net electric field is in the y-direction, it has only a y-component. The magnitude of the net electric field at the point is: \( E_{net} = E_y = 4.30 × 10^6 N/C \) The direction of the net electric field can be determined by the charges on the wires. Since the wires have opposite charges, they exert attractive forces on each other, and their electric fields point inward (toward each other). Hence, the direction of the net electric field at the point is parallel or along with the plane containing the two wires. So, the magnitude of the electric field at the point midway between the two wires and 40 cm above the plane containing them is \(4.30 × 10^6 N/C\), and its direction is parallel or along the plane containing the two wires.