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**Hint:**We know that the electric dipole direction is from negative charge to the positive one. We then find the direction of the electric field on the equatorial line. Then find the angle between them.

Formulae Used:

$ p = qd $

Where, $ p $ is the dipole moment of the dipole, $ q $ is the magnitude of the either charges and $ d $ is the distance between the charges.

**Complete Step By Step Solution**

Here, we take $ O $ as the midpoint of a dipole $ AB $

Thus, $ AO = OB = d/2 $

Here, $ P $ is a point on the equatorial line.

Now, Let, $ OP = x $

So, $ AP = BP = {\left[ {{{\left( {d/2} \right)}^2} + {x^2}} \right]^{1/2}} $

Now, Electric field strength on $ P $ due to $ + q $ is,

$ {E_{ + q}} = kq/{\left( {AP} \right)^2} $

Where, k is the universal electric force constant.

Again, Electric field strength on $ P $ due to $ -q $ is,

$ {E_{ - q}} = kq/{\left( {BP} \right)^2} $

Now, For the resultant electric field we need to take the component of both the fields in the direction of the resultant and then add them,

Thus, Resultant Field,

$ {E_r} = {E_{ + q}}cos\theta {\text{ }} + {E_{ - q}}cos\theta $

By putting all the values, we get,

$ {E_r} = 2kqcos\theta /{\left[ {{{\left( {d/2} \right)}^2} + {x^2}} \right]^{1/2}} $

Now, From our knowledge of fields , it is clear that the direction of the resultant or the net electric field on the equatorial line is opposite to the direction of the dipole.

**Thus, the angle between them is $ {180^0} $ which is (C).**

**Note**

The direction of the electric field is towards a negative charge and away from a positive charge. The distance between the dipole and the point $ P $ on the equatorial line is a variable, thus, this analogy could be extended till infinity.

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