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Potential on Axial Line of Dipole

Potential on the axis of a dipole (θ = 0° or 180°). Positive on the +q side, negative on the −q side. Follows from the general expression with cos 0° = 1.
Class 11Class 12
Derivation

From the general formula

General dipole potential: V=pcos⁡θ4πε0r2V = \dfrac{p\cos\theta}{4\pi\varepsilon_0 r^2}

On the axial line, θ=0°\theta = 0° (on the +q+q side) or θ=180°\theta = 180° (on the −q-q side).

+q+q side (θ=0°\theta = 0°, cos⁡θ=1\cos\theta = 1):

Vaxial=+p4πε0r2V_{axial} = +\frac{p}{4\pi\varepsilon_0 r^2}

−q-q side (θ=180°\theta = 180°, cos⁡θ=−1\cos\theta = -1):

Vaxial=−p4πε0r2V_{axial} = -\frac{p}{4\pi\varepsilon_0 r^2} Vaxial=±p4πε0r2\boxed{V_{axial} = \pm\frac{p}{4\pi\varepsilon_0 r^2}}

Direct verification

For point P at distance rr on the +q+q side (distances: r+=r−ar_+ = r - a, r−=r+ar_- = r + a):

V=q4πε0(1r−a−1r+a)=q4πε0⋅2ar2−a2≈p4πε0r2V = \frac{q}{4\pi\varepsilon_0}\left(\frac{1}{r-a} - \frac{1}{r+a}\right) = \frac{q}{4\pi\varepsilon_0}\cdot\frac{2a}{r^2 - a^2} \approx \frac{p}{4\pi\varepsilon_0 r^2}

Consistent with the general formula.

Remember
Compare with the axial field: $E_{axial} = 2p/4\pi\varepsilon_0 r^3$. Note $E = -dV/dr$ gives $d/dr(p/4\pi\varepsilon_0 r^2) = -2p/4\pi\varepsilon_0 r^3$, consistent with $E = -dV/dr$.