The figure can be drawn according to given conditions.
Where, in the above figure at the point X the charge is +q, at the point Y the charge is −2q and at the point Z the charge is +q.
The distances are given as,
XY=YZ=a YP=r PX=r+a PZ=r−a
The electrostatic potential at point P is given as,
V= 1 4π ε 0 [ q XP − 2q YP + q ZP ]
By substituting the given values in the above expression, we get
V= 1 4π ε 0 [ q r+a − 2q r + q r−a ] = q 4π ε 0 [ r( r−a )−2( r+a )( r−a )+r( r+a ) r( r+a )( r−a ) ] = q 4π ε 0 [ r 2 −ra−2 r 2 +2 a 2 + r 2 +ra r( r 2 − a 2 ) ] = 2q a 2 4π ε 0 r 3 ( 1− a 2 r 2 )
Since, the value of r a ≫1, So a r ≪1 and the term a 2 r 2 can be neglected.
The value of electric potential is given as,
V= 2q a 2 4π ε 0 r 3
Thus, the expression of electrical potential is V∝ 1 r 3 , for a dipole is V∝ 1 r 2 and for a monopole system is V∝ 1 r .