A short dipole of moment $p = 1 \times 10^{-9}$ C·m. The electric potential due to it at 0.3 m from the centre on a line making 60° with the dipole axis is (use $1/4\pi\varepsilon_0 = 9\times10^9$ N·m²·C⁻²):

For a short dipole at distance $r$, angle $\theta$ from the axis, $V = (1/4\pi\varepsilon_0)\, p\cos\theta / r^2$. Substituting: $V = (9\times10^9)(1\times10^{-9})\cos 60°/(0.3)^2 = 9 \times 0.5/0.09 = 50$ V. On axis ($\theta=0$) the value would be 100 V; on the equator ($\theta=90°$) it would be 0.