A cube of side b has a charge q at seven of its vertices. The electric field due to this charge distribution at the centre of this cube will be:

If same charge is placed at all eight vertices then net electric field is zero i.e., vector sum of all eight fields is zero.E→1+E→2+E3→+E4→+E5→+E6→+E7→+E8→=0 E1→+E→2+E→3+E→4+E→5+E→6+E→7=E→8=4k⋅q3b2As half of the diagonal of the cube is r=3b2Hence when seven charges are at vertices then electric field at the center =4k⋅q3b2