A proton with a speed of 2.0 x 10⁵ m/s falls through a potential difference V and thereby increases its speed to 4.0 × 10⁵ m/s.Through what potential difference did the proton fall?

Question

Quizplus · Accepted Answer

The change in kinetic energy of the proton is equal to the work done by the electric field, which is the charge of the proton times the potential difference. Using the formula ΔKE = qV, where ΔKE = 0.5 * m * (v_f^2 - v_i^2), and knowing the charge of a proton is approximately 1.6 x 10^-19 C, we can solve for V to find it is approximately 630 V.

A Proton with a Speed of 2