# finding the intial speed closest to x=0?

Two large conducting parallel plates A and B are separated by 2.4 m. A uniform field of 1500 V/m, in the positive x-direction, is produced by charges on the plates. The center plane at x = 0.0 m is an equipotential surface on which V = 0. An electron is projected from x = 0.0 m, with an initial kinetic energy K = 300 eV, in the positive x-direction as shown.

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- DambarudharLv 78 years agoFavorite Answer
K = 300 eV = 300*{1.602*10^(- 19)} J

=> (1/2)mu² = 300*{1.602*10^(- 19)} J

where m = mass of the electron = 9.108 * 10^(- 31) kg

u = speed of the electron at x = 0.0 m

u = √(2K/m) =√[ 2*300{1.602*10^(- 19)} / {9.108 * 10^(- 31)} ]

= √[{(2*3*1.602)/9.108} * 10^14] m/s = 1.027*10^7 m/s

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