A particle has a charge of +1.5µC and moves from point Ato point B, a distance o

Question

A particle has a charge of +1.5µC and moves from point Ato point B, a distance of 0.19m. The particle experiences aconstant electric force, and its motion is along the line of actionof the force. The difference between the particle's electricpotentail energy at A and B is EPEA - EPEB =+6.00*10-4 J. a) Find the magnitude and direction of the electric force thatacts on the particle. b) Find the magnitude and direction of the electric field thatthe particle experences. This is a question from the Cutnell 7th edition book, Chaper19 question 4. Thanks and PLEASE HELP! A particle has a charge of +1.5µC and moves from point Ato point B, a distance of 0.19m. The particle experiences aconstant electric force, and its motion is along the line of actionof the force. The difference between the particle's electricpotentail energy at A and B is EPEA - EPEB =+6.00*10-4 J. a) Find the magnitude and direction of the electric force thatacts on the particle. b) Find the magnitude and direction of the electric field thatthe particle experences. This is a question from the Cutnell 7th edition book, Chaper19 question 4. Thanks and PLEASE HELP!

Explanation / Answer

   from the theory we can see that the workdone bythe electric force as the particle moves from point A to point B isgiven    by    WAB = EPEA -EPEB    if s is the displacement and F is the forcethen    WAB = F s (direction of motion isfrom A to B)    the electric field is given by    E = F / qo (a)    so the electric force will be    F = (EPEA - EPEB) /s       = +6.00 x 10-4J / 0.19 m       = ........ N (b)    the electric field will be    E = F / qo       = ......... N / c (thedirection is from A to B)       = ......... N / c (thedirection is from A to B)

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