An electron is launched from rest at point toward point

PHYSICS · Multi-part · AP · Physics 2: Algebra-based · Electric Force Field and Potential · Electric Force Field and Potential

Problem

An electron is launched from rest at point AA toward point BB, which lies 5.0 cm5.0\,\text{cm} away along the axis of a uniform electric field. Point BB is at an electric potential ΔV=+200 V\Delta V=+200\,\text{V} relative to point AA. Assume the only force on the electron is the electrostatic force. Use e=1.60×10−19 Ce=1.60\times10^{-19}\,\text{C} and me=9.11×10−31 kgm_e=9.11\times10^{-31}\,\text{kg}. (a) Calculate the speed of the electron as it reaches point BB. (b) The initial launch speed is now changed to 3.0×106 m ⁣⋅ ⁣s−13.0\times10^{6}\,\text{m}\!\cdot\!\text{s}^{-1} while the potential difference remains +200 V+200\,\text{V}. Determine the potential difference (magnitude and sign) that would be required, instead of +200 V+200\,\text{V}, to bring an electron with that initial speed to rest exactly at point BB.

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