An electron is near a positive ion of charge +9e and a negative ion of charge 8e

Question

An electron is near a positive ion of charge +9e and a negative ion of charge 8e (see the figure below). (Take a = 3.93 µm, b = 3.64 µm, and = 61.6°.)

(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

direction

  °

(b) Find the magnitude and direction of the electron's instantaneous acceleration (Let right be the +x-direction. Measure the angle counter-clockwise from the +x-axis.)

magnitude

direction

  °

magnitude

direction

  °

Explanation / Answer

A)

Force on electron due to -8e charge:

F1 = k*(8e)/(3*10^-6)^2 directed away from -8e

Similarly, force on it due to +9e:

F2 = k*(9e)/(3.93*10^-6)^2 directed towards +9e

So, net force =( (k*9e/(3.93*10^-6)^2)*cos(61.6 deg) - k*8e/(3*10^-6)^2)^2 + (k*9e/(3.93*10^-6)^2)*sin(61.2 deg)

= 7.97*10^11*9*10^9*1.6*10^-19*1.6*10^-19

= 1.83*10^-16 N<------ magnitude

Direction = 61.6 + atan(5.11/6.11) = -101.5 deg

= 360-101.5 = 258.5 deg

B)

Acceleration =(1.83*10^-16) /(9.1*10^-31)

= 2.02*10^14 m/s2 <------ magnitude

Direction = 258.5 deg

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