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Question: Two electric dipoles A, B with respective dipole moments \({\vec d_A} = - 4qa\hat i\) and \({\vec d_...

Two electric dipoles A, B with respective dipole moments d⃗A=−4qai^{\vec d_A} = - 4qa\hat i and d⃗B=−2qai^{\vec d_B} = - 2qa\hat i are placed on the x-axis with a separation R. The distance from A at which both of them produce the same potential is
A. 2R2+1\dfrac{{\sqrt 2 R}}{{\sqrt 2 + 1}}
B. R2+1\dfrac{R}{{\sqrt2 + 1}}
C. 2R2−1\dfrac{{\sqrt 2 R}}{{\sqrt 2 - 1}}
D. R2−1\dfrac{R}{{\sqrt2 - 1}}

Explanation

Solution

Hint
The potential due to a dipole depends directly on the dipole moment and inversely on the distance. This potential acts in a direction towards the dipole from the point at which it is being calculated.
⇒V=kPcos⁡θx2\Rightarrow V = k\dfrac{{P\cos \theta }}{{{x^2}}}, where V is the electric potential at a distance x from the dipole moment P. θ\theta is the angle between the distance vector and the electric potential towards the dipoles. K is constant given as 14πε0\dfrac{1}{{4\pi {\varepsilon _0}}}.

Complete step by step answer
We have two electric dipoles on a straight line separated by a distance R. We are asked to find the distance from one of the dipoles at which the electric potential produced by the two dipoles would be equal.
The information provided to us is:
Dipole moment of A d⃗A=−4qai^{\vec d_A} = - 4qa\hat i
Dipole moment of B d⃗B=−2qai^{\vec d_B} = - 2qa\hat i
Distance between A and B: R
Let us assume the point at which the electric potential is the same to lie at a distance of x from A. This means the distance of this point from B will be R−xR - x. Now, we know that the electric potential has a direction towards the dipoles from the point. Hence from point x, the potential due to point A and point B will be in opposite directions as it lies in between.
Now, we only need to compare their magnitude which is given as:
⇒V=kPx2\Rightarrow V = k\dfrac{P}{{{x^2}}}
According to the question,
⇒VA=VB\Rightarrow {V_A} = {V_B}
⇒kPAx2=kPB(R−x)2\Rightarrow k\dfrac{{{P_A}}}{{{x^2}}} = k\dfrac{{{P_B}}}{{{{(R - x)}^2}}}
Substituting the values in this, we get:
⇒−4qax2=−2qa(R−x)2\Rightarrow \dfrac{{ - 4qa}}{{{x^2}}} = \dfrac{{ - 2qa}}{{{{(R - x)}^2}}}
⇒2x2=1(R−x)2\Rightarrow \dfrac{2}{{{x^2}}} = \dfrac{1}{{{{(R - x)}^2}}}
Solving for x, we get:
⇒2(R−x)2=x2\Rightarrow 2{(R - x)^2} = {x^2}
⇒x=±2(R−x)\Rightarrow x = \pm \sqrt 2 (R - x)
We have two solutions for the value of x. First let’s consider the positive one:
⇒x=2(R−x)=2R−2x\Rightarrow x = \sqrt 2 (R - x) = \sqrt 2 R - \sqrt 2 x
⇒x+2x=2R\Rightarrow x + \sqrt 2 x = \sqrt 2 R
Solving further:
⇒x(1+2)=2R\Rightarrow x(1 + \sqrt 2 ) = \sqrt 2 R
⇒x=2R(1+2)\Rightarrow x = \dfrac{{\sqrt 2 R}}{{(1 + \sqrt 2 )}}
Similarly, solving for the negative part, we get:
⇒x=−2(R−x)=−2R+2x\Rightarrow x = - \sqrt 2 (R - x) = - \sqrt 2 R + \sqrt 2 x
⇒2x−x=2R\Rightarrow \sqrt 2 x - x = \sqrt 2 R
Solving further:
⇒x(2−1)=2R\Rightarrow x(\sqrt 2 - 1) = \sqrt 2 R
⇒x=2R(2−1)\Rightarrow x = \dfrac{{\sqrt 2 R}}{{(\sqrt 2 - 1)}}
Hence, the options (A) and (C) are both correct.

Note
Just like any other potential, electric potential is the amount of work needed to move an electric charge. This parameter is important to take account of because we are always looking for convenient sources of energy. If one charge moves in the field of another, a potential would be generated that would ultimately produce some energy. That’s why knowing how potentials work is important.