Calculating Charge in an Earth Electric Field: Gauss' Law

[carus88]

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 Vm1. 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 Vm1. 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?

 

[Redbelly98]

... Use Gauss’ law ...

Use Gauss' law, not Coulomb's law.

 

[merryjman]

This is a pretty standard and basic Gauss's Law problem. I would suggest finding an introductory physics textbook and reading the appropriate section. You might even find this same problem solved, or something like it.