Question

A proton is released from rest at the origin in a uniform electric field that is directed in the positive x 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 = 2.5 \text{ m}x=2.5 m

Answer 1

Explanation:

It is given that,

Electric field, $E=950\ N/C$

We need to find the change in the electric potential energy of the proton-field system when the proton travels to x = 2.5 m

From the conservation of energy, the loss in potential energy is equal to the gain in kinetic energy and kinetic energy is is equal to the work done.

$W=F\times x$

$W=q\times E\times x$

$W=1.6\times 10^{-19}\times 950\times 2.5$

$W=3.8\times 10^{-16}\ J$

So, the change in electric potential energy of the proton field system is $3.8\times 10^{-16}\ J$. Hence, this is the required solution.