Two parallel infinite line charges with linear charge densities $+\lambda \, C/m$ and $-\lambda \, C/m$ are placed at a distance of 2R in free space. What is the electric field mid-way between the two line charges ?

Question

There are two cubical Gaussian surface carrying charges as shown. Find ratio of fluxes through surface $C_1$ and $C_2$:

Answer 1

To find the electric field midway between two parallel infinite line charges with linear charge densities +\lambda \, \text{C/m} and -\lambda \, \text{C/m}, we begin by visualizing the setup. The line charges are placed at a distance of 2R apart in free space.

Since both are infinite line charges, by Gauss's Law, the electric field due to a single infinite line charge in free space can be expressed as:

E = \frac{\lambda}{2\pi\epsilon_0 r}

where \lambda is the linear charge density, r is the radial distance from the line charge, and \epsilon_0 is the permittivity of free space.

The electric field due to the positive line charge at midpoint (distance R from the charge) is directed away from the line charge:

E_+ = \frac{\lambda}{2\pi\epsilon_0 R}

The electric field due to the negative line charge at midpoint (distance R from the charge) is directed towards the line charge, hence in the same direction as E_+ since midway point lies between them:

E_- = \frac{\lambda}{2\pi\epsilon_0 R}

Since both fields are in the same direction at the midpoint, the net electric field at the midpoint is the sum of these two fields:

E_{\text{net}} = E_+ + E_- = \frac{\lambda}{2\pi\epsilon_0 R} + \frac{\lambda}{2\pi\epsilon_0 R} = \frac{\lambda}{\pi\epsilon_0 R}

Thus, the net electric field midway between the two line charges is:

\frac{\lambda}{\pi\epsilon_0 R} N/C

This correctly matches the given answer.

Answer 2