Courses for Kids
Free study material
Offline Centres
Store Icon

Two parallel infinite line charges $ + \lambda $ and $ - \lambda $ are placed with a separation distance R in free space. The net electric field exactly midway between the two line charges is
(A) Zero
(B) $\dfrac{{2\lambda }}{{\pi {E_0}R}}$
(C) $\dfrac{\lambda }{{\pi {E_0}R}}$
(D) $\dfrac{1}{{2\pi {E_0}R}}$

Last updated date: 18th Jun 2024
Total views: 53.1k
Views today: 1.53k
53.1k+ views
Hint: The direction can be identified by seeing the distance or separation between the charges as: At exactly midway so the distance can be exactly half so here, the parallel infinite charges both positive and negative charges mean in the same direction.

Formula used:
Different formulas will be used to solve the problem which is mentioned below as:
$ {E_1} = \dfrac{\lambda }{{2\pi {E_0}\dfrac{R}{2}}} \\
  {E_2} = \dfrac{{ - \lambda }}{{2\pi {E_0}\dfrac{R}{2}}} \\
Where R is distance between the separation
${E_0}$ Is epsilon value
E is the electrical field
$\lambda $ Is the positive and negative both are the infinite line charge

Complete Step by step answer:
As we know that a point charge is a hypothetical charge located at a single point in space.
And then the electric field is a vector. There are multiple point charges present. The net electric field at any point is the vector sum of the electric fields due to the individual charges.
By this image we can understand that separation R so as it’s at its exactly midway so we can take $\dfrac{R}{2}$ as their mid separation and this both the charges are in same direction
As parallel is given so both the direction will be same let be left to right
  ${E_1} = \dfrac{\lambda }{{2\pi {E_0}\dfrac{R}{2}}}$ and ${E_2} = \dfrac{{ - \lambda }}{{2\pi {E_0}\dfrac{R}{2}}}$
So, the net energy is,
${E_{net}} = {E_1} + {E_2}$
${E_{net}} = \dfrac{\lambda }{{2\pi {E_0}\dfrac{R}{2}}} + \dfrac{{ - \lambda }}{{2\pi {E_0}\dfrac{R}{2}}}$
$ \Rightarrow {E_{net}} = \dfrac{{2\lambda }}{{\pi {E_0}R}}$ So we get,
Hence the net electric field is: ${E_{net}} = \dfrac{{2\lambda }}{{\pi {E_0}R}}$

Hence the correct option is B that is $\dfrac{{2\lambda }}{{\pi {E_0}R}}$.

In question probably we get the hint so first we need to think about the direction of the charges. And then what is the distance of separation between the charges.
So first basically the electric field of an individual and then to get a total combining electric fields of the charges.