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

The change in potential energy is  

Explanation:

From the question we are told that

     The  magnitude of the uniform electric field  is  

      The  distance traveled by the electron is  

Generally the force on this electron is  mathematically represented as

Where F is the force and  q is the charge on the electron which is  a constant value of  

    Thus  

Generally the work energy theorem can be mathematically represented as

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

Also  workdone on the electron can also  be represented as

Where   considering that the movement of the electron is along the x-axis  

        So

substituting values

Now From the law of energy conservation

Where is the change  in  potential energy  

Thus  

54 degrees incident angle = reflected angle

angle is from normal turned to line of ray