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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.
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