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Gauss's law

  1. Aug 16, 2008 #1
    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?
     
  2. jcsd
  3. Aug 16, 2008 #2

    Redbelly98

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    Use Gauss' law, not Coulomb's law.
     
  4. Aug 16, 2008 #3
    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.
     
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