Potential on a cone

All six 60 degree slices contribute equally. So calculate the potential due to the entire (full) cone, and then multiply that by $\frac{5}{6}$ . The key to this problem is knowing what the area element is. It can be derived using standard multi-variate calculus as indicated here . Suppose the cone has height $H$ and radius $R$ , and that its length is along the $z$ -direction.

$\large{dS = \sqrt{1 + \frac{H^2}{R^2}} \, r \, dr \, d \theta}$

The differential charge is thus:

$\large{dQ = \sigma \, dS = \sigma \, \sqrt{1 + \frac{H^2}{R^2}} \, r \, dr \, d \theta}$

The relationship between the vertical position and the radial position is:

$\large{z = H - \frac{H}{R} \, r}$

The distance between an area element and the top of the cone is:

$D = \sqrt{r^2 + (H-z)^2} = \sqrt{r^2 + \frac{H^2}{R^2} \, r^2} = \sqrt{1 + \frac{H^2}{R^2}} \, r$

The differential electric potential contribution is:

$\large{dU = \frac{dQ}{4 \, \pi \, \epsilon_0 \,D} = \frac{\sigma \, \sqrt{1 + \frac{H^2}{R^2}} \, r \, dr \, d \theta}{4 \, \pi \, \epsilon_0 \, \sqrt{1 + \frac{H^2}{R^2}} \, r} \\ = \frac{\sigma \, dr \, d \theta}{4 \, \pi \, \epsilon_0} }$

The total electric potential for the full cone is:

$\large{U = \int_0^{2 \pi} \int_0^R \frac{\sigma \, dr \, d \theta}{4 \, \pi \, \epsilon_0} = \frac{\sigma R}{2 \epsilon_0}}$

The electric potential for the reduced cone is:

$\large{U_r = \frac{5}{6} \, U = \frac{5 \sigma R}{12 \epsilon_0}}$