A solid non-conducting cube of side l and uniformly distributed charge q, has electric field E and potential V at one of its vertex, imagine this cube to be made of 8 smaller cubes of side l/2. if one such cube is removed, find the new field and potential at the point where the vertex used to be.
2. The attempt at a solution
Intuitively i thought, if E is field due to a cube of side l and charge q, the smaller cube has side l/2 and charge q/8. usually electric field is proportional to q and inversely proportional to r, assuming the same we get the field due to smaller cube E/2 which seems fair since it was much closer than the other cubes. by the same logic, potential usually falls of as 1/r, so potential due to smaller cube is V/4 but this looks a bit fishy...can someone show me the right way to look at this? I'll be really grateful