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

Respuesta :

Answer:

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

Explanation:

From the question we are told that

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

      The  distance traveled by the electron is  [tex]x = 2.50 \ m[/tex]

Generally the force on this electron is  mathematically represented as

     [tex]F = qE[/tex]

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

    Thus  

      [tex]F = 950 * 1.60 **10^{-19}[/tex]

      [tex]F = 1.52 *10^{-16} \ N[/tex]

Generally the work energy theorem can be mathematically represented as

          [tex]W = \Delta KE[/tex]

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

Also  workdone on the electron can also  be represented as

        [tex]W = F* x *cos( \theta )[/tex]

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

        So

             [tex]\Delta KE = F * x cos (0)[/tex]

substituting values

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

          [tex]\Delta KE = 3.8*10^{-16} J[/tex]

Now From the law of energy conservation

       [tex]\Delta PE = - \Delta KE[/tex]

Where [tex]\Delta PE[/tex] is the change  in  potential energy  

Thus  

        [tex]\Delta PE = - 3.8*10^{-16} \ J[/tex]