Electrostatic potential at a point due to an electric dipole: Consider two equal and opposite charges separated by a small distance 2a. The point P is located at a distance r from the midpoint of the dipole. Let `theta` be the angle between the line OP and dipole axis AB.
Let `r_(1)` be the distance of the point P from +q and `r_(2)` be the distance of point P from -q.
Potential at P due to charge +q `=(1)/(4pi epsi_(0)) q/r_(1)`
Potential at P due to charge -q `=-(1)/(4pi epsi_(0)) q/r_(2)`
Total potential at the point P, `V=(1)/(4pi epsi_(0)) q (1/r_(1)-1/r_(2))` .....(1)
Suppose if the point P is far away from the dipole, such that `r gt gt a`, the equation can be expressed in terms of r. By the cosine law for triangle BOP,
`r_(1)^(2)=r^(2)+a^(2)-2ra cos theta=r^(2) (1+a^(2)/r^(2)-(2a)/(r) cos theta)`
Since the point P is very far from dipole, then `r gt gt a`. As a result the term `a^(2)/r^(2)` is very small and can be neglected. Therefore
`r_(1)^(2)=r^(2) (1-2a"" (cos theta)/(r))`
`1/r_(1)=1/r (1-(2a)/(r) cos theta)`
since `a/r lt lt 1`, we can use binomial theorem and retain the terms up to first order
`1/r_(1)=1r/r (1+a/r cos theta)`
Similarly applying the cousine law for triangle, AOP,
`r_(2)^(2)=r^(2)+a^(2)-2ra cos (180-theta)" Since cos"(180-theta)=-cos theta" we get " r_(2)^(2)=r^(2)+a^(2)+2ra cos theta`
Neglecting the term `a^(2)/r^(2) ("because "r gt gt a)`
`r_(2)^(2) =r^(2) (1+(2a cos theta)/(r))`
Using Binomial theorem, we get
`1/r_(2)=1/r (1-a"" (cos theta)/(r)) ..........(3)`
Substituting equations (3) and (2) in equaltion (1)
`V=(1)=(1)/(4pi epsi_(0)) q[1/r (1+a (cos theta)/r)-1/4 (1-a (cos theta)/(r))]=(q)/(4piepsi_(0)) [1/r(1+a (cos theta)/(r)-1 +a(cos theta)/(r))]`
`V=(1)/(4pi epsi_(0)) (2aq)/(r^(2)) cos theta`
But the electric dipole moment p=2qa and we get,
`V=(1)/(4pi epsi_(0)) ((p cos theta)/(r^(2)))`
Now we can write p cos `theta=vecp, hatr`, is the unit vector from hte point O to point P. Hence the electric potential at a point P due to an electric dipole is given by `V=(1)/(4piepsi_(0)) (vecp. hatr)/(r^(2)) (r gt gt a) ...........(4)`
Equation (4) is the valid for distance very large compared to the size of the dipole. But for a point dipole, the equation (4) is valid for any distance.
Special cases:
Case (I) If the point P lies on the axial line of the dipole on the side of +q, then `theta=0`. Then the electric potential becomes `V=(1)/(4pi epsi_(0)) p/r^(2)`
Case (ii) If the point P lies on the axial line of the dipole on the side of -q then `theta=180^(@)`, them `V=- (1)/(4pi epsi_(0)) p/r^(2)`
Case (iii) If the point P lies on the equatorial line of the dipole, thena `theta=90^(@)`. Hence V=0
