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Question: Two parallel infinite line charges \( + \lambda \) and \( - \lambda \) are placed with a separation ...

Two parallel infinite line charges +λ+ \lambda and λ- \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) 2λπE0R\dfrac{{2\lambda }}{{\pi {E_0}R}}
(C) λπE0R\dfrac{\lambda }{{\pi {E_0}R}}
(D) 12πE0R\dfrac{1}{{2\pi {E_0}R}}

Explanation

Solution

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:
E1=λ2πE0R2 E2=λ2πE0R2 {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
E0{E_0} Is epsilon value
E is the electrical field
λ\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 R2\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
E1=λ2πE0R2{E_1} = \dfrac{\lambda }{{2\pi {E_0}\dfrac{R}{2}}} and E2=λ2πE0R2{E_2} = \dfrac{{ - \lambda }}{{2\pi {E_0}\dfrac{R}{2}}}
So, the net energy is,
Enet=E1+E2{E_{net}} = {E_1} + {E_2}
Enet=λ2πE0R2+λ2πE0R2{E_{net}} = \dfrac{\lambda }{{2\pi {E_0}\dfrac{R}{2}}} + \dfrac{{ - \lambda }}{{2\pi {E_0}\dfrac{R}{2}}}
Enet=2λπE0R\Rightarrow {E_{net}} = \dfrac{{2\lambda }}{{\pi {E_0}R}} So we get,
Hence the net electric field is: Enet=2λπE0R{E_{net}} = \dfrac{{2\lambda }}{{\pi {E_0}R}}

Hence the correct option is B that is 2λπE0R\dfrac{{2\lambda }}{{\pi {E_0}R}}.

Note:
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.