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) $\dfrac{1}{{2\pi {E_0}R}}$
Answer
299.1k+ 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 $\dfrac{{2\lambda }}{{\pi {E_0}R}}$.
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.
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 $\dfrac{{2\lambda }}{{\pi {E_0}R}}$.
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.
Recently Updated Pages
If the magnetizing field on a ferromagnetic material class 12 physics JEE_Main

A point charge is placed at the corner of a cube The class 12 physics JEE_Main

What changes occur if the monochromatic light used class 12 physics JEE_Main

The unit of specific conductance is A Ohm B Ohmmetre class 12 physics JEE_Main

Which lens is used in magnifying glass A Concave lens class 12 physics JEE_Main

Sir C V Raman won the Nobel Prize in which year A 1928 class 12 physics JEE_Main

Trending doubts
JEE Main 2026: Exam Dates, Session 2 Updates, City Slip, Admit Card & Latest News

Understanding the Electric Field of a Uniformly Charged Ring

Understanding Atomic Structure for Beginners

Electron Gain Enthalpy and Electron Affinity Explained

Derivation of Equation of Trajectory Explained for Students

How to Convert a Galvanometer into an Ammeter or Voltmeter

Other Pages
CBSE Class 12 Physics Question Paper 2026: Download SET-wise PDF with Answer Key & Analysis

JEE Advanced Percentile vs Marks 2026: JEE Main Cutoff, AIR & IIT Admission Guide

What Are Current and Potential Difference in Electricity?

Understanding Uniform Acceleration in Physics

Hybridisation in Chemistry – Concept, Types & Applications

Understanding Geostationary and Geosynchronous Satellites

