A proton with a speed of 2.0 x $10 ^ { 5 }$ m/s accelerates through a potential difference and thereby increases its speed to 4.0 x $10 ^ { 5 }$ m/s. Through what magnitude potential difference did the proton accelerate? (e = 1.60 × 10^-19 C , m_proton = 1.67 × 10^-27 kg)

Question

Quizplus · Accepted Answer

The change in kinetic energy equals the work done by the electric field: $ \Delta KE = q \Delta V $. The change in kinetic energy is $ \frac{1}{2} m (v_f^2 - v_i^2) $. Solving for $ \Delta V $, we get $ \Delta V = \frac{m (v_f^2 - v_i^2)}{2q} $. Substituting the given values, $ \Delta V = \frac{1.67 \times 10^{-27} \times ((4.0 \times 10^5)^2 - (2.0 \times 10^5)^2)}{2 \times 1.60 \times 10^{-19}} \approx 630 $ V.

A Proton with a Speed of 2 10510 ^ { 5 }105

A Proton with a Speed of 2 $10 ^ { 5 }$