Physics

Physics

Problem

An electron starting from rest is accelerated through a potential difference of 1200 V1200\,\text{V}. Determine its speed after the acceleration. \newline (Take electron charge e=1.6×10−19 Ce = 1.6\times10^{-19}\,\text{C} and mass me=9.1×10−31 kgm_e = 9.1\times10^{-31}\,\text{kg}.)
Intuitive explanation

The key is recognizing that electrical potential energy converts directly into kinetic energy—no other forces do work. Set the work done by the electric field, eVeV, equal to the kinetic energy gained, 12mev2\frac{1}{2}m_ev^2, and solve for vv. This energy approach is much cleaner than trying to find acceleration and distance separately.

Intuitive explanation

The key is recognizing that electrical potential energy converts directly into kinetic energy—no other forces do work. Set the work done by the electric field, eVeV, equal to the kinetic energy gained, 12mev2\frac{1}{2}m_ev^2, and solve for vv. This energy approach is much cleaner than trying to find acceleration and distance separately.

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