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Torque on rod with electric field & proton

  1. May 26, 2008 #1
    Reading ahead. Don't understand components. Question: A uniform electric field is exerting a force of 3.2X10^-18 Nt u on a proton located at the west end of an east/west directed horizontal rod. r(proton)=-0.5 m e. The torque (r X F) exerted on the rod caused by the interaction of this field and the proton is: Torque = ( ) e + ( ) n + ( ) u.
  2. jcsd
  3. May 27, 2008 #2
    Using cross product, I get the following: Torque = 0 e + 1.6*10^-18 n + 0 u

    (is this correct?)
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