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

]]>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. ]]>