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Torque on rod with electric field & proton

  • #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.
 

Answers and Replies

  • #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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