Question

The electric field at an axial point of the short dipole is $\mathop {{E_1}}\limits^ \to$if the electric field at equatorial point of the same dipole is $\mathop {{E_2}}\limits^ \to$, then which of the following is correct.
A) $\mathop {{E_1}}\limits^ \to . \mathop {{E_2}}\limits^ \to < 0$
B) $\mathop {{E_1}}\limits^ \to . \mathop {{E_2}}\limits^ \to > 0$
C) $\mathop {{E_1}}\limits^ \to . \mathop {{E_2}}\limits^ \to = 0$
D) $\mathop {{E_1}}\limits^ \to + \mathop {{E_2}}\limits^ \to = 0$

Answer 1

Hint
here, first of all we use the formula of electric field at axial and equatorial point i.e.
Formula of electric field at axial point is $\mathop {{E_1}}\limits^ \to = \dfrac{{2KP}}{{{r^3}}}$
Formula of electric field at equatorial point is $\mathop {{E_2}}\limits^ \to = \dfrac{{KP}}{{{r^3}}}$, the using both formula we can check which option is correct.

Complete step by step answer
Let the electric field at axial point is $\mathop {{E_1}}\limits^ \to$ and the electric field at equatorial point is$\mathop {{E_2}}\limits^ \to$
As, here we want to find the relation between the electric field at axial and equatorial point then we will use the formulas of electric field at both the points.
Now, electric field at axial point is $\mathop {{E_1}}\limits^ \to = \dfrac{{2KP}}{{{r^3}}}$ …………………… (1)
Where, P is the dipole moment
$K = \dfrac{1}{{4\pi { \in _0}}}$, $ε_0$ is the permittivity of the free space.
R is the distance of the point.
Similarly, the electric field at the equatorial point is $\mathop {{E_2}}\limits^ \to = \dfrac{{KP}}{{{r^3}}}$ ………………….. (2)
Where, P is the dipole moment
$K = \dfrac{1}{{4\pi { \in _0}}}$, $ε_0$ is the permittivity of the free space.
R is the distance of the point.
Now, in order to find the relation between both electric fields, we observe from the equation (1) and (2), the electric field at the axial point is twice the electric field at the equatorial point. i.e. $\mathop {{E_1}}\limits^ \to = 2\mathop {{E_2}}\limits^ \to$
Noe, as from equation (1), $\mathop {{E_1}}\limits^ \to > 0$
And from equation (2), $\mathop {{E_2}}\limits^ \to > 0$
Therefore, the product of both the electric fields is greater than zero i.e. $\mathop {{E_1}}\limits^ \to . \mathop {{E_2}}\limits^ \to > 0$
Hence, option (B) is correct.

Note
Axial line is the line which is passing through the positive and negative charges and the point lies on that line is called the axial point.
Equatorial line is the perpendicular line to the line passing through the positive and negative charges and the point lies on that line is known as the equatorial point.
Here, it must be notice that the electric field at the axial point due to electric dipole is $E = \dfrac{q}{{4\pi { \in _0}}}\left[ {\dfrac{{4ra}}{{{{\left( {{r^2} - {a^2}} \right)}^2}}}} \right]$, a is the distance between the two charges. But as here it is given that short dipole, therefore we consider $a < < < r$ and hence above formula will become $\mathop {{E_1}}\limits^ \to = \dfrac{{2KP}}{{{r^3}}}$, $P = q \times 2a$.
Similarly, electric field at equatorial point is $E = \dfrac{{2qa}}{{4\pi { \in _0}{{\left( {{r^2} + {a^2}} \right)}^{3/2}}}}$, then again for short dipole$a < < < r$, hence formula will become $\mathop {{E_2}}\limits^ \to = \dfrac{{KP}}{{{r^3}}}$. Where symbols have the same meaning as explained in the above question.