Question

A proton is released from rest in a uniform electric field whose magnitude is 5000 V/m. Through what potential difference will it have passed after moving 0.25 meters? How fast will it be going
after it has traveled 0.25 meters?

Answer 1

Since the potential difference is equal to the product of the field intensity by the distance traveled by the charged particle, then $U=E\times d=5000 \times 0.25= 1250$ V.

Since $U\times q=E_c=\frac{m\times V^2}{2}$ , then

$V=\sqrt{\frac{2\times U\times q}{m}}$ $=\sqrt{\frac{2\times 1250\times 1.6\times 10^{-19}}{1.67\times 10^{-27}}}\approx 4.89\times 10^5$ m/s.