A proton is released from rest in a uniform electric field of magnitude 50000 v/m directed along the positive x axis and undergoes a displacement of 0.5 m in the direction of the electric field as shown. find the amount of the kinetic energy gained after it has moved 0.5 m..
For the law of conservation of energy, the amount of kinetic energy the proton gained in [tex]d=0.5 m[/tex] is equal to the amount of electric potential energy it losts covering the same distance.
The potential difference across which the proton travelled is given by [tex]\Delta V = E d[/tex] where E is the electric field intensity. Replacing the numbers, we get [tex]\Delta V = (50000 V/m)(0.5 m)=25000 V[/tex]
The electric potential energy lost by the proton is given by [tex]\Delta U = q \Delta V[/tex] where [tex]q=1.6 \cdot 10^{-19}C[/tex] is the charge of the proton. Therefore, this quantity is equal to [tex]\Delta U = (1.6 \cdot 10^{-19}C)(25000 V)=4 \cdot 10^{-15}J[/tex]
And based on what we said at the beginning, this electric potential energy lost by the proton is exactly equal to the amount of kinetic energy it gained: [tex]\Delta K = 4 \cdot 10^{-15}J[/tex]
