An electron of mass 9.11 x 10^-31 kg has an initial speed of 3.40 x 10^5 m/s. It

Question

An electron of mass 9.11 x 10^-31 kg has an initial speed of 3.40 x 10^5 m/s. It travels in a straight line, and its speed increases to 6.80 x 10^5 m/s in a distance of 5.60 cm. Assume its acceleration is constant. Determine the magnitude of the force exerted on the electron. Your response is within 10% of the correct value. This may be due to roundoff error, or you could have a mistake in your calculation. Carry out all intermediate results to at least four-digit accuracy to minimize roundoff error. N Compare this force (F) with the weight of the electron (F_g), which we ignored.

Explanation / Answer

The acceleration comes from v^2 = v0^2 + 2*a*x

so a = (v^2 - v0^2)/(2*x) = ((6.80x10^5)^2 - (3.40x10^5)^2)/(2*0.056) = 3.09x10^12 m/s^2

a) So F = m*a =9.11e-31*3.09e12= 2.82e-18 N and

b) since weight = m*g =9.11e-31*9.81=8.94e-30 N

then F/Fg = (m*a)/(m*g) = a/g = 3.09x10^12/ 9.81= 3.15x10^11

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