No real need to use the 3 Unit maths approach. You would only use 3 unit integration if they were going to ask you to do things like:
- find the distance the electron travelled,
- find the time it took for the electron to leave the electric field and so on.
a) The energy gained from the acceleration in the electron gun is transformed into Kinetic Energy.
E = qV
E = 1.6 x 10^{-19} x 1000
= 1.6 x 10^{-16} J = ½ mv^{2}
Thus v = √ (2 x 1.6 x 10^{-16}/9.1 x ^{-31})
v = 1.87 x 10^{7}m/s
b) Now, the electron enters the Electric field of strength 10000 N/C.
The horizontal veolcity has no effect on this calculation. It would be like throwing a ball off a cliff with a certain velocity. The acceleration due to gravity still acts in a verticle direction unaffected by the horizontal motion.
F = ma = Eq
9.1 x ^{-31} a = 10000 x 1.6 x 10^{-19}
Thus
a = 1.76 x 10^{15}