Question

Two parallel infinite line charges with linear charge densities \(+λ \frac Cm\) and \(-λ \frac Cm\) are placed at a distance of \(2R\) in free space. What is the electric field mid-way between the two line charges

• A. zero
• B. \(\frac{2\lambda}{\pi\epsilon_0R}\)\(N/C\)
• C. \(\frac{\lambda}{\pi\epsilon_0R}\)\(N/C\)
• D. \(\frac{\lambda}{2\pi\epsilon_0R}\)\(N/C\)

Answer 1

The Correct Option is C

To determine the electric field mid-way between two parallel infinite line charges with linear charge densities \(+λ\) and \(-λ\), which are placed at a distance of \(2R\) apart in free space, we follow these steps:

1. Consider the two line charges placed along a line, separated by \(2R\). Since a line charge produces an electric field radially outward (or inward for negative charge) and perpendicular to the line, we need to calculate the resultant electric field at the midpoint of the two charges.
2. The position of the midpoint is \(\frac{R}{2}\) from each line charge. The electric field due to an infinite line charge at a distance \(r\) is given by the formula: E = \frac{\lambda}{2\pi\epsilon_0r} where \(\lambda\) is the linear charge density, and \(\epsilon_0\) is the permittivity of free space.
3. At the midpoint, the distance to each line charge is \(R\), therefore the electric field due to each at this point is: E = \frac{\lambda}{2\pi\epsilon_0R}
4. Because the two fields are in opposite directions (due to the opposite charges), they add up linearly in magnitude at the midpoint (i.e., their vectors point in the same direction):
5. The total electric field at the midpoint is: E_{\text{total}} = \frac{\lambda}{2\pi\epsilon_0R} + \frac{\lambda}{2\pi\epsilon_0R}
6. Thus, E_{\text{total}} = \frac{\lambda}{\pi\epsilon_0R}
7. This electric field is directed from the positive line charge towards the negative line charge, perpendicular to the line joining the charges.

Hence, the correct answer is \(\frac{\lambda}{\pi\epsilon_0R}\) \(N/C\).