Explanation:
It is given that,
Electric field, [tex]E=950\ N/C[/tex]
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.
[tex]W=F\times x[/tex]
[tex]W=q\times E\times x[/tex]
[tex]W=1.6\times 10^{-19}\times 950\times 2.5[/tex]
[tex]W=3.8\times 10^{-16}\ J[/tex]
So, the change in electric potential energy of the proton field system is [tex]3.8\times 10^{-16}\ J[/tex]. Hence, this is the required solution.
