An electron is projected with an initial velocity v_o = 3.0 times 10^7 m/s along

Question

An electron is projected with an initial velocity v_o = 3.0 times 10^7 m/s along the y-axis, which is the center line between a pair of charged plates. The plates are 1.00 m long and are separated by 0.10 m. A uniform electric field E, in the positive x-direction, is presented between the plates. The magnitude of the acceleration of the electron is 4.3 times 10^15 m/s^2. In Fig. 21.9, the magnitude of the electric field between the plates is closest to: A) 19,000 N/C B) 25,000 N/C C) 14,000 N/C D) 22,000 N/C E) 17,000 N/C

Explanation / Answer

By definition, E = F/ q; since the particle at hand is an electron, E = F/e. Of course, F =ma, so E = ma/e, or E = a / (e/m).

e/m, electron's charge-to-mass ratio is known since 1897, and was first measured by J. J. Thompson, even before electron mass could be ascertained. e/m = 1.759 × 10^11 C/kg. Substituting this in the above expression,

E = 4.3 E15 N/kg / 1.759 E11 C/kg = 2.445 E4 N/C =25000 N/C

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