An electron is projected with an initial velocity o = 3.4 x 107 m/s along the y-

Question

An electron is projected with an initial velocity o = 3.4 x 107 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 8.9 x 1015 m/s2. In Fig. 21.9, the magnitude of the electric field is changed so that an electron, projected with the same initial velocity, barely clears the far edge of one of the plates. The magnitude of this electric field is closest to: 900 N/C 580 N/C 820 N/C 660 N/C 690 N/C

Explanation / Answer

force applied = 9.1 * 10^-31 * 8.9 * 10^15 = 8.1 * 10^-15 N thus the electric field = F/q = 50625 N/C now when the new electric field is applied the distance in the x-direction = 0.05 m in y direction = 1 m now the motion would be a part of circular motion thus the force applied = mv^2/r where r = 1 m thus force = 9.1 * 10^-31 * (3.4 * 10^7) = 1.05 * 10^-15 thus the E = 656.25 N/C thus the answer should be 660 N/C

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