Find out the magnitude of  electric field intencity at point (2,0,0), due to a dipole of dipole moment , P_vector = i_cap + 3 j_cap kept at origin. also find out the potential at that point 

Hi !  this question can be answered in a very intiuitive way...

Consider the dipole to be a superposition of 2 independent dipoles having p vectors as follows

p₁ = i^{^} and p₂ = 3j^{^}

Observe that the point(2,0,0) lies on the x axis..thus for dipole 1 it is like an axial point and for dipole 2 it is like an equatorial point...now use the formulas ...

E₁ = 2kp/r³   = 2x9x10⁹/8 i^{^} = k/4 i^{^}

E₂ = -kp/r³   = -9x10⁹/8 i^{^} = -k/8j^{^}

Thus E = E₁ + E₂ = (k/4 i^{^} + -k/8j^{^} ) NC^-1

Just check the calculations..rest all is fine I feel

Similarly for potential at the point you can use the formula V = k(p.r^{^})/r²

Thus V = k( (i^{^}+3j^{^} ). i^{^})/ 4 = k/4 volt.