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

Answer 1

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  

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