Question

Find out the magnitude of electric field intensity and electric potential due to a dipole of dipole moment $\overrightarrow P = \hat i + \sqrt 3 \hat j$ kept at origin at following points.
(i) $\left( {2,0,0} \right)$
(ii) $\left( { - 1,\sqrt 3, 0} \right)$

Answer 1

Hint:You can solve this question rather easily by trying to recall the formulae of electric field intensity and electric potential due to a dipole, which are as followed:
$E = \dfrac{{kp}}{{{r^3}}}\sqrt {1 + 3{{\cos }^2}\theta } $ and,
$V = \dfrac{{k\left( {\overrightarrow p .\overrightarrow r } \right)}}{{{r^2}}}$

Complete step by step answer:

We will be trying to solve the question just like we explained in the hint section of the solution to this question.
Firstly, we can see that the value of $\tan \theta $ for the given dipole $\overrightarrow P = \hat i + \sqrt 3 \hat j$ is $\sqrt 3 $, hence, this makes an angle of ${60^ \circ }$ with the X-axis.
The point given in (i) is at X-axis, hence, it makes an angle of ${60^ \circ }$ with it too.
Now, we can write:
$E = \dfrac{{kp}}{{{r^3}}}\sqrt {1 + 3{{\cos }^2}\theta } $,
Where, $\theta = {60^ \circ }$
$p = \left| {\overrightarrow P } \right| = 2$
And $r = 2$ as well
The equation becomes:
$E = \dfrac{{2k}}{{{2^3}}}\sqrt {1 + 3\left( {\dfrac{1}{4}} \right)} $
Further solving, we get:
$E = \dfrac{k}{{{2^2}}}\sqrt {\dfrac{7}{4}} = \dfrac{{\sqrt 7 k}}{8}$
Similarly, for electric potential, we can write:
$V = \dfrac{{k\left( {\overrightarrow p .\overrightarrow r } \right)}}{{{r^2}}}$
And then substitute the values that we have been given in the question:
$V = \dfrac{{k\left( {\hat i + \sqrt 3 \hat j} \right).\left( {2\hat i} \right)}}{{{2^2}}}$
Further solving, we get:
$V = \dfrac{{2k}}{4} = \dfrac{k}{2}$
Following the same steps for the case (ii) and just replacing the value of
$\overrightarrow r = - \hat i + \sqrt 3 \hat j$ and solving the same equation as above, we get:
\[V = \dfrac{{k\left( {\hat i + \sqrt 3 \hat j} \right).\left( { - \hat i + \sqrt 3 \hat j} \right)}}{{{2^2}}}\]
Further solving, we get:
$V = \dfrac{{2k}}{{{2^2}}} = \dfrac{k}{2}$

Note:Many students makes do not consider the step of finding the angle between the dipole and the given point as important and thus, only imply the formula of either of the two of finding electric field intensity at either the pole or the axis, which is a completely wrong approach to find the solution of the given question.