QuestionDetails: Each of the protons in a particle beam has a kinetic energy of3

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QuestionDetails: Each of the protons in a particle beam has a kinetic energy of3.25 x 10-15 J. What are the magnitude and direction ofthe electric field that will stop these protons in a distance of1.25 m? QuestionDetails: Each of the protons in a particle beam has a kinetic energy of3.25 x 10-15 J. What are the magnitude and direction ofthe electric field that will stop these protons in a distance of1.25 m? QuestionDetails: Each of the protons in a particle beam has a kinetic energy of3.25 x 10-15 J. What are the magnitude and direction ofthe electric field that will stop these protons in a distance of1.25 m?

Explanation / Answer

We know that :               Work done W = F * d = (1/2)mv2 But  F = E q    But  F = E q          Hence   W=   qE * d=   (1/2)mv2                           Electric field is :                        E = mv2 / 2qd                                                 =[ (3.25 x 10-15 J) / 2(1.602 x10-19 C * 1.25 m) ]                           = ------------- N / C Hope this helps u!

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