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 solve the problem of finding the electric field mid-way between two parallel infinite line charges with linear charge densities \(+λ \frac{C}{m}\) and \(-λ \frac{C}{m}\), we need to understand the field contribution from each line charge:

1. The electric field due to an infinite line charge with linear charge density \(\lambda\) at a perpendicular distance \(r\) from the line is given by the formula:

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

where \(\epsilon_0\) is the permittivity of free space.

2. Consider the midpoint between the two lines as the point of interest. Since the line charges are separated by a distance \(2R\), the midpoint will be at a distance \(R\) from both line charges.
3. At the midpoint, the electric fields due to both line charges will be equal in magnitude but opposite in direction because one line has a positive linear charge density and the other a negative charge density.
4. Thus, the electric field due to the line with \(\lambda\) at the midpoint \(E_1\) is: E_1 = \frac{\lambda}{2\pi\epsilon_0 R}.
5. Similarly, the electric field due to the line with \(-\lambda\) at the midpoint \(E_2\) is: E_2 = \frac{\lambda}{2\pi\epsilon_0 R}.
6. These two fields \(E_1\) and \(E_2\) are in the same direction (since they are opposite charges), so they add together to give the resultant electric field.

Thus, the resultant electric field \(E\) at the midpoint is:

E = E_1 + E_2 = \frac{\lambda}{2\pi\epsilon_0 R} + \frac{\lambda}{2\pi\epsilon_0 R} = \frac{\lambda}{\pi\epsilon_0 R}.

Therefore, the electric field mid-way between the two line charges is:

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

The correct answer is thus the option: \(\frac{\lambda}{\pi\epsilon_0 R}\) N/C.