1 v d ? 2 v d ? 3 v ? 2 d 4 v ? d 5 v d? 6 v ? 2 d 7

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1. V d ǫ 0 2. V d ǫ 0 3. V ǫ 0 2 d 4. V ǫ 0 d 5. V 0 6. V ǫ 0 2 d 7. V ǫ 0 d correct 8. V 0 Explanation: Once again, we use a gaussian pillbox; this time, we place the surface S so that it encloses part of the inner surface of the right plate. The field between the plates is vector E = ( V/d x since the figure indicates the left plate is at a higher potential. Applying Gauss’ law, V d A = σ ǫ 0 A σ = V ǫ 0 d .
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  • Spring '08
  • Turner
  • Electrostatics, Magnetic Field, Electric charge, A. Gauss

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