Question

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..

Answer 1

For the law of conservation of energy, the amount of kinetic energy the proton gained in $d=0.5 m$ 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
$\Delta V = E d$
where E is the electric field intensity. Replacing the numbers, we get
$\Delta V = (50000 V/m)(0.5 m)=25000 V$

The electric potential energy lost by the proton is given by
$\Delta U = q \Delta V$
where $q=1.6 \cdot 10^{-19}C$ is the charge of the proton. Therefore, this quantity is equal to
$\Delta U = (1.6 \cdot 10^{-19}C)(25000 V)=4 \cdot 10^{-15}J$

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:
$\Delta K = 4 \cdot 10^{-15}J$