Question

Two infinitely long straight wires having linear charge densities \(-\lambda\) and \(3\lambda\) are held vertically parallel to each other, distance \(r\) apart in free space. Find the nature and magnitude of the force/length exerted by one wire on the other.

Hint:

Like charges or distributions with identical signs repel each other, whereas unlike or oppositely charged bodies attract. The electric field of a long straight line charge falls off slowly as \(\frac{1}{r}\).

Answer 1

Concept: According to Gauss's Law, the magnitude of the electric field \(\vec{E}\) produced by an infinitely long straight wire carrying a uniform linear charge density \(\lambda\) at a perpendicular radial distance \(r\) from it is given by: \[ E = \frac{\lambda}{2\pi\varepsilon_0 r} \] The electrostatic force \(\vec{F}\) acting on a charge element \(dq\) placed in an electric field \(\vec{E}\) is \(\vec{F} = dq \vec{E}\). Thus, the force per unit length on a second parallel wire carrying linear charge density \(\lambda'\) is given by \(f = \lambda' E\).

Step 1: Determine the nature of the force.
Analyze the signs of the linear charge densities.
The first wire has a uniform negative linear charge density of \(-\lambda\), while the second wire possesses a positive linear charge density of \(+3\lambda\). Since the two charge distributions carry opposite signs, they will experience a mutually attractive force pulling them toward one another. Therefore, the nature of the force is completely attractive.

Step 2: Calculate the magnitude of the electric field produced by the first wire.
Apply Gauss's law formula for a line charge distribution.
The electric field magnitude \(E_1\) generated by the wire with linear charge density \(-\lambda\) at a perpendicular separation distance \(r\) is: \[ E_1 = \frac{\lambda}{2\pi\varepsilon_0 r} \]

Step 3: Determine the force per unit length on the second wire.
Multiply the linear charge density of the target wire by the external electric field.
The force per unit length (\(f\)) experienced by the second wire containing a linear charge density of magnitude \(3\lambda\) due to the presence of field \(E_1\) is: \[ f = (3\lambda) \times E_1 \] Substituting our mathematical expression for \(E_1\): \[ f = 3\lambda \left( \frac{\lambda}{2\pi\varepsilon_0 r} \right) = \frac{3\lambda^2}{2\pi\varepsilon_0 r} \] Thus, the magnitude of the force per unit length is exactly \(\frac{3\lambda^2}{2\pi\varepsilon_0 r}\).