An electron with a speed of 6.39 times 10^8 cm/s in the positive direction of an

Question

An electron with a speed of 6.39 times 10^8 cm/s in the positive direction of an x axis enters an electric field of magnitude 1.21 times 10^3 N/C, traveling along a field line in the direction that retards its motion. (a) How far will the electron travel in the field before stopping momentarily, and (b) how much time will have elapsed? (c) If the region containing the electric field is 7.04 mm long (too short for the electron to stop within it), what fraction of the electron's initial kinetic energy will be lost in that region? (a) Number .096 Units b) Number 3.007e-8 Units c) Units .0606 Units

Explanation / Answer

c)

m = mass of electron = 9.1 x 10-31 kg

v = speed of electron = 6.39 x 106 m/s

kinetic energy of electron is given as

KEi = (0.5) m v2 = (0.5) (9.1 x 10-31) (6.39 x 106)2 = 1.86 x 10-17 J

L = length of region = 7.04 mm = 0.00704 m

E = electric field = 1210 N/C

V = potential difference = E L = 1210 x 0.00704 = 8.52 volts

q = charge on electron = 1.6 x 10-19

energy lost = qV = (1.6 x 10-19) (8.52) = 1.4 x 10-18

fraction : (1.4 x 10-18)/(1.86 x 10-17 ) = 0.075

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