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)

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The kinetic energy change equals the work done by the electric field: $ \Delta KE = q \Delta V $. Calculate the change in kinetic energy: $ \Delta KE = \frac{1}{2} m (v_f^2 - v_i^2) $. Substitute values: $ \Delta KE = \frac{1}{2} \times 1.67 \times 10^{-27} \times ((4.0 \times 10^5)^2 - (2.0 \times 10^5)^2) $. Solve for $ \Delta V $: $ \Delta V = \frac{\Delta KE}{e} \approx 630 $ V.

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

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