Question

An electron is placed at rest in a uniform electric field of 227N/C, 270.0 degree next to the negatively charged plate. What is the final velocity of the electron if it has a mass of 9.11E-31 kg and accelerates for 0.074m ?

Answer 1

The change of energy is equal to work done.

$$\frac{mv^2}{2}=qEd$$

$$v=\sqrt{2qEd/m}$$

$$v=\sqrt{\frac{2\times 1.6\times 10^{-19}\times 227\times 0.074}{9.11\times10^{-31}}}=2.43\times 10^6\:\rm m/s$$