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

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)

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 times the potential difference. Using the formula: $ \Delta KE = \frac{1}{2} m (v_f^2 - v_i^2) = e \Delta V $, we find $ \Delta V = \frac{1}{2} \times 1.67 \times 10^{-27} \times ((4.0 \times 10^5)^2 - (2.0 \times 10^5)^2) / 1.60 \times 10^{-19} \approx 630 $ V.

A Proton with a Speed of 2 10510^5105 M/s Accelerates Through a Potential Difference and Thereby Increases Its

A Proton with a Speed of 2 $10^5$ M/s Accelerates Through a Potential Difference and Thereby Increases Its