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Question
Derive the expression for the electric potential due to an electric dipole at a point on its axial line.
Solution
i) Let P be the point at which electric potential is required.
Potential at P due to − q charge,
`V_1=-q/(4piepsilon_0r_1)`
Potential at P due to + q charge,
`V_2=q/(4piepsilon_0r_2)`
Potential at P due to the dipole,
`V=V_1+V_2`
`V=q/(4piepsilon_0)[1/r_2-1/r_1]` .....(i)
Now, by geometry
`r_1^2=r^2+a^2+2arcostheta`
`r_2^2=r^2+a^2+2arcos(180^@-theta)`
`r_2^2=r^2+a^2-2arcostheta`
`r_1^2=r^2(1+a^2/r^2+(2a)/rcostheta)`
If r >> a, `a/r` is small. Therefore,`a^2/r^2` can be neglected.
`r_1^2=r^2(1+(2a)/rcostheta)`
`1/r_1=1/r(1+(2a)/rcostheta)^(1/2)`
similarly, `1/r_2=1/r(1-(2a)/rcostheta)^(1/2)`
Putting these values in (i), we obtain
`V=q/(4piepsilon_0)[1/r(1-(2a)/rcostheta)^(1/2)-1/r(1+(2a)/rcostheta)^(1/2)]`
Using Binomial theorem and retaining terms up to the first order in`a/r`, we obtain
`V=q/(4piepsilon_0r)[1+a/rcostheta-(1-a/rcostheta)]`
`V=q/(4piepsilon_0r)[1+a/rcostheta-1+a/rcostheta]`
`V=q/(4piepsilon_0r)[(2a)/rcostheta]`
`V=(qxx2acostheta)/(4piepsilon_0r^2)`
`V=(Pcostheta)/(4piepsilon_0r^2)`
For axial line put θ=0°
`V=(Pcos(0))/(4piepsilon_0r^2)`
`V=P/(4piepsilon_0r^2)`
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