A proton (m = 1.7 *10 ^- ²⁷ kg,q = +1.6 *10 ^- ¹⁹ C)starts from rest at point A and has a speed of 40 km/s at point B.Only electric forces act on it during this motion.Determine the electric potential difference V_B - V_A. A)+8.5 V B)-8.5 V C)-4.8 V D)+4.8 V E)-17 V

Question

Quizplus · Accepted Answer

The change in kinetic energy equals the work done by the electric field, which is equal to the charge times the potential difference. Using the formula $ \Delta KE = q \Delta V $, and substituting the given values, the potential difference is calculated to be -8.5 V.

A Proton (M = 1