Question

derivation of electric potential at a point on axial line of an electric dipole

Answer 1

Suppose P is a point on the axial position of dipole 

length of dipole = 2a 

suppose point P is the distance "r" from the center  of the dipole

 we know that

$V = \frac{1}{4 \pi E_0} . \frac{Q}{r}$

So , potential at P due to to "q" is 

 $V_q = \frac{1}{4 \pi E_0} . \frac{q}{(a+r)}$

potential at P due to "-q" is

$V_-q = \frac{1}{4 \pi E_0} . \frac{-q}{(r-a)}$

Total potential at P is 

[tex]V = V_q + V_-q [/tex]

$V = \frac{1}{4 \pi E_0} . \frac{Q}{(a+r)} + \frac{1}{4 \pi E_0} . \frac{-q}{(r-a)}$

$V = \frac{q}{4 \pi E_0} [ \frac{1}{a+r} + \frac{1}{a-r} ]$

$V = \frac{q}{4 \pi E_0} . \frac{2a}{a^2-r^2}$

Answer 2

The NCERT formula is correct if used with vector notation as the direction of electric dipole moment is from -q(negative charge) to +q(positive charge) and the resultant electric field at any point on equatorial line is in direction opposite to the electric dipole moment so negative sign is used. The other formula which does not include negative sign is applicable to calculate the magnitude(not direction) of electric field.