Question

An electron is near a positive ion of charge +9e and a negative ion of charge-8e (see the figure below). (Take a = 3.95 μm, b-3.60 μm , and θ-60.6.) -8e +9e Electron (a) Find the magnitude and direction of the resultant force on the electron. (Let right be the +x-direction. Measure the angle counter-clockwise from the +x-axis.) magnitude 1.78e-16N direction 90 (b) Find the magnitude and direction of the electrons instantaneous acceleration (Let right be the +x-direction. Measure the angle counter-clockwise from the +x-axis.) magnitude 1.95e14 m/s*2 direction 90

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Answer #1

The pictorial depiction of the problem is, sino E cose +9e The x component of the force is, Fos+Ford = heigcos (-i), kg: (9x1The direction of the force is, tan - 1.784x10-15 -3.326x10-15 @=-28.20° (b) The acceleration of the electron is, m 1.78x10-4

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