The earth’s electric field is measured at a height of 200 m and is found to be directed vertically downwards and to have a strength of 100 Vm1. At a height of 300 m, the direction of the field is found to be the same, but the field’s strength has decreased to 60 Vm1. Use Gauss’ law to determine the amount of charge contained in a cube of side 100 m with one face parallel to the earth’s surface located at a height between 200 m and 300 m. (You may ignore the curvature of the earth.) THIS IS FOR A PHYSICS DEGREE MODULE RESIT. I HAVE DRAWN RELEVANT DIAGRAMS YET I HAVE NO IDEA WHERE TO START. aNY GUIDANCE IN THE RIGHT DIRECTION WOULD BE A MASSIVE HELP. do I use coulombs law and work out the respective charges?