Then -3V_0/4a

]]>That’s the almost the right answer, but it is in the -z direction so you need a minus sign…

]]>Although it makes more sense that Luke Barnes is right. You didn’t post “nope” after that comment.

]]>No. That’s a very wrong argument!. In fact in the neighbourhood of the origin the equipotential is in the equatorial plane, perpendicular to the direction you claim. By symmetry the electric field runs vertically through the origin, which tells you that the potential must be proportional to z there.

]]>Ok, I found an error in my calculation. New guess is 3V/(4a).

]]>So, I try a smart way: there is a path in the equipotential hemispheres (without electric field), and along the diameter of the sphere; the field of the sphere is symmetric, so that the electric field have the direction of the diameter (that connect the centres of the upper and lower hemisphere), and it is constant because of the Gauss’s law (there is not transversal flux).

If the distance between the Hemisphere is little, then the charge distribution don’t change with the distance between hemispheres, and the electric field multiplied by the distance tends to zero in the insulated equator.

I suppose that the field along the diameter is V_0/(2a), because the line integral of the electric field is zero along a closed path.

If it so, then the problem use all the electrostatic laws. ]]>