An electron is released from rest at the negative plate of a parallel plate capa

Question

An electron is released from rest at the negative plate of a parallel plate capacitor. The charge per unit area on each plate is ? = 1.20 10-7 C/m2, and the plates are separated by a distance of 1.0 10-2 m. How fast is the electron moving just before it reaches the positive plate?
m/s

Explanation / Answer

Electric field between the plates of the capacitor = s/eo N/C Force acting on charge q = qs/eo N Work done on moving this charge across distance ?x = q(s/eo)?x This work gets converted into kinetic energy of the charge having mass m => (1/2) mv^2 = q(s/eo)?x => v = v[(2q(s/eo)?x)/m] ... [As stated by you] Now, plugging values, mass of electron, m = 9.10938188 × 10-31 kg and charge of electron, q = 1.60217646 × 10-19 C => velocity, v = v[2 * (1.60217646 × 10-19) * (1.2 x 10^-7 / 8.854 x 10-12) * (1 x 10^-2 / 9.10938188 × 10-31] m/s

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