
Two charges 9e and 3e are placed at a separation r. The distance of the point where the electric field intensity will be zero, is:
$\eqalign{
& {\text{A) }}\dfrac{r}{{1 + \sqrt 3 }}{\text{from 9e charge}} \cr
& {\text{B) }}\dfrac{r}{{1 + \sqrt {\dfrac{1}{3}} }}{\text{from 9e charge}} \cr
& {\text{C) }}\dfrac{r}{{1 - \sqrt 3 }}{\text{from 3e charge}} \cr
& {\text{D) }}\dfrac{r}{{1 + \sqrt {\dfrac{1}{3}} }}{\text{from 3e charge}} \cr} $
Answer
141.3k+ views
Hint: Here in this question we need to find the neutral point , where the force experienced by a unit positive charge is zero , A neutral point is a point at which the resultant magnetic field is zero and it is obtained when horizontal component of earth’s field is balanced by the produced magnet.
Complete step by step solution:
Two charges 9e and 3e separation=r
Let the distance at which electric field intensity from 9e charge = 0, be x and from 3e charge it will be r-x.
Due to a system of two like point charge, the electric field of both the charges at neutral point will be equal
$\eqalign{
& \Rightarrow \dfrac{1}{{4\Pi {\varepsilon _o}}} \times \dfrac{{9e \times 1}}{{{a^2}}} = \dfrac{1}{{4\Pi {\varepsilon _o}}} \times \dfrac{{3e}}{{{{(r - x)}^2}}} \cr
& \Rightarrow 3{(r - x)^2} = {x^2} \cr
& {\text{solving we get,}} \cr
& \Rightarrow \sqrt 3 (r - x) = x \cr
& or,\sqrt 3 r - \sqrt 3 x = x \cr
& \Rightarrow \sqrt 3 r = x(1 + \sqrt 3 ) \cr
& \therefore x = \dfrac{{r\sqrt 3 }}{{1 + \sqrt 3 }}{\text{ from 9e charge}} \cr} $
Hence, the option B is right.
Additional Information: The electric field at a given distance from a point charge is a vector amount, which points away from the positive charge and towards the negative charge. Its magnitude follows the inverse square law: it is proportional to the charge and inversely proportional to the distance.
The zero field location has to be on a line running between two point charges because it is the only place where field vectors can point in exactly opposite directions. This cannot occur between two opposite charges because field vectors from both charges point towards negative charge. It is to be on one side or the other of the two, where the vectors point in opposite directions.
Note: If both charges are the same, then the resulting potential is not zero at any finite point because such charges will have the same sign on the potentials and can therefore never be added to zero so such a point is only at infinity.
Complete step by step solution:
Two charges 9e and 3e separation=r
Let the distance at which electric field intensity from 9e charge = 0, be x and from 3e charge it will be r-x.
Due to a system of two like point charge, the electric field of both the charges at neutral point will be equal
$\eqalign{
& \Rightarrow \dfrac{1}{{4\Pi {\varepsilon _o}}} \times \dfrac{{9e \times 1}}{{{a^2}}} = \dfrac{1}{{4\Pi {\varepsilon _o}}} \times \dfrac{{3e}}{{{{(r - x)}^2}}} \cr
& \Rightarrow 3{(r - x)^2} = {x^2} \cr
& {\text{solving we get,}} \cr
& \Rightarrow \sqrt 3 (r - x) = x \cr
& or,\sqrt 3 r - \sqrt 3 x = x \cr
& \Rightarrow \sqrt 3 r = x(1 + \sqrt 3 ) \cr
& \therefore x = \dfrac{{r\sqrt 3 }}{{1 + \sqrt 3 }}{\text{ from 9e charge}} \cr} $
Hence, the option B is right.
Additional Information: The electric field at a given distance from a point charge is a vector amount, which points away from the positive charge and towards the negative charge. Its magnitude follows the inverse square law: it is proportional to the charge and inversely proportional to the distance.
The zero field location has to be on a line running between two point charges because it is the only place where field vectors can point in exactly opposite directions. This cannot occur between two opposite charges because field vectors from both charges point towards negative charge. It is to be on one side or the other of the two, where the vectors point in opposite directions.
Note: If both charges are the same, then the resulting potential is not zero at any finite point because such charges will have the same sign on the potentials and can therefore never be added to zero so such a point is only at infinity.
Recently Updated Pages
Young's Double Slit Experiment Step by Step Derivation

Difference Between Circuit Switching and Packet Switching

Difference Between Mass and Weight

JEE Main Participating Colleges 2024 - A Complete List of Top Colleges

JEE Main Maths Paper Pattern 2025 – Marking, Sections & Tips

Sign up for JEE Main 2025 Live Classes - Vedantu

Trending doubts
Learn About Angle Of Deviation In Prism: JEE Main Physics 2025

Formula for number of images formed by two plane mirrors class 12 physics JEE_Main

Electron Gain Enthalpy and Electron Affinity for JEE

Physics Average Value and RMS Value JEE Main 2025

Explain the construction and working of a GeigerMuller class 12 physics JEE_Main

Atomic Structure - Electrons, Protons, Neutrons and Atomic Models

Other Pages
Collision - Important Concepts and Tips for JEE

Introduction to Dimensions With Different Units and Formula for JEE

The perfect formula used for calculating induced emf class 12 physics JEE_Main

Displacement-Time Graph and Velocity-Time Graph for JEE

A planoconvex lens f20cm is silvered at the plane surface class 12 physics JEE_Main

The ratio of the lengths densities masses and resistivities class 12 physics JEE_Main
