There are different ways to do this.
One of the ways is to divide the sphere into N sectors of equal width, like segments of an orange. Each of the N pieces will have equal surface area.
Another way is to divide the sphere into N sectors of equal width, and then to divide along the equator, so that each will be a half-segment. Each of the 2*N pieces will have equal surface area.
Another way is to divide the sphere into N partial sectors of equal width, like chopping the ends off of a sphere and then dividing the rest into N equal partial sectors. If you do this at the correct angle, then each of the partial sectors will have the same surface area as each of the pole caps. If you measure angles from the "left" as 0 and designate the starting angle as t and end at angle Pi-t, then you can achieve this when equality of surface areas when t = arccos(N/(N+2)) and each of the (N+2) pieces will have surface area 4*r^2*Pi/(N+2) . The two pole caps are the same shape as each other, and the wedges are the same shape as the pole caps, but the pole caps and the wedges are not the same shape as each other.
This could obviously be extended by dividing along the equator and also dividing the pole caps into 2. You can keep generalizing to get sectors of the pole caps and partial sectors of the main part.
Now, what you cannot do is cover the surface of the sphere with circular-edged slivers all of the same size and shape.