Q.The potential due to a dipole at axial point is

\dfrac{1}{4\pi \epsilon_0} \dfrac{2p}{r^2}

\dfrac{1}{4\pi \epsilon_0} \dfrac{p}{r^2}

\dfrac{1}{4\pi \epsilon_0} \dfrac{p}{r}

\dfrac{1}{4\pi \epsilon_0} \dfrac{2p}{r^3}

The correct solution involves applying the fundamental concept to derive the final value step by step...

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