Question

Two long straight thin wires AB and CD have linear charge densities \( 10 \, \text{mC/m} \) and \( -20 \, \text{mC/m} \), respectively. They are kept parallel to each other at a distance of 1 m. Find the magnitude and direction of the net electric field at a point midway between them.

Hint:

For two parallel charged wires, the net electric field at a point between them is the vector sum of the individual fields. The fields due to opposite charges act in opposite directions.

Answer 1

The electric field due to a long straight charged wire with a linear charge density \( \lambda \) at a distance \( r \) from it is given by the formula:
\[ E = \frac{\lambda}{2\pi\epsilon_0 r} \] where:
\( \lambda_1 = 10 \, \mu\text{C/m} = 10 \times 10^{-6} \, \text{C/m} \) (charge density of wire AB),
\( \lambda_2 = -20 \, \mu\text{C/m} = -20 \times 10^{-6} \, \text{C/m} \) (charge density of wire CD),
\( r = 0.5 \, \text{m} \) (distance from the point to each wire), \( \epsilon_0 = 8.85 \times 10^{-12} \, \text{C}^2 / \text{N m}^2 \) is the permittivity of free space.
The net electric field at the midpoint between the two wires is the vector sum of the fields due to each wire. Let’s calculate the electric fields \( E_1 \) and \( E_2 \) at the midpoint:
\[ E_1 = \frac{\lambda_1}{2\pi \epsilon_0 r} = \frac{10 \times 10^{-6}}{2\pi (8.85 \times 10^{-12}) (0.5)} = 3.59 \times 10^6 \, \text{N/C} \] \[ E_2 = \frac{\lambda_2}{2\pi \epsilon_0 r} = \frac{-20 \times 10^{-6}}{2\pi (8.85 \times 10^{-12}) (0.5)} = -7.18 \times 10^6 \, \text{N/C} \] Step 1: Calculate the net electric field The direction of the electric fields due to both wires:
The electric field due to wire AB (with \( \lambda_1 \)) points away from the wire, toward wire CD.
The electric field due to wire CD (with \( \lambda_2 \)) points toward the wire, so it also points toward wire AB.
Since both fields point toward wire CD, the net electric field is the sum of the magnitudes:
\[ E_{\text{net}} = E_1 + E_2 = 3.59 \times 10^6 + 7.18 \times 10^6 = 10.77 \times 10^6 \, \text{N/C} \] Thus, the magnitude of the net electric field is:
\[ E_{\text{net}} = 10.77 \times 10^6 \, \text{N/C} \] and the direction is toward wire CD (left). % Correct Answer Correct Answer:}
The magnitude of the net electric field is \( 10.77 \times 10^6 \, \text{N/C} \). The direction of the net electric field is toward wire CD (left).