A proton is released from rest at the origin in a uniform electric field that is directed in the positive xx direction with magnitude 950 \t

Question

A proton is released from rest at the origin in a uniform electric field that is directed in the positive xx direction with magnitude 950 \text{ N/C}950 N/C. What is the change in the electric potential energy of the proton-field system when the proton travels to x

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Khang Minh 3 months 2021-08-18T07:15:07+00:00 1 Answers 28 views 0

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    2021-08-18T07:16:46+00:00

    Answer:

    The change in potential energy is  \Delta  PE =  -  3.8*10^{-16} \ J

    Explanation:

    From the question we are told that

         The  magnitude of the uniform electric field  is  E =  950 \ N/C

          The  distance traveled by the electron is  x =  2.50 \ m

    Generally the force on this electron is  mathematically represented as

         F =  qE

    Where F is the force and  q is the charge on the electron which is  a constant value of  q =  1.60*10^{-19} \ C

        Thus  

          F  =  950  * 1.60 **10^{-19}

          F  = 1.52 *10^{-16} \ N

    Generally the work energy theorem can be mathematically represented as

              W =  \Delta  KE

    Where W is the workdone on the electron by the  Electric field and  \Delta  KE  is the change in kinetic energy

    Also  workdone on the electron can also  be represented as

            W =  F* x  *cos(  \theta )

    Where  \theta  =  0 ^o considering that the movement of the electron is along the x-axis  

            So

                 \Delta  KE  =  F  * x  cos  (0)

    substituting values

             \Delta  KE  =  1.52 *10^{-16}  * 2.50   cos  (0)

              \Delta  KE   =  3.8*10^{-16} J

    Now From the law of energy conservation

           \Delta PE  =  -  \Delta  KE

    Where \Delta  PE is the change  in  potential energy  

    Thus  

            \Delta  PE =  -  3.8*10^{-16} \ J

                   

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