Question

what is the speed of an electron released at the negative plate before reaching the positive plate where distance is 0.02 m potential difference is 250 volt and electric field is 125,000 N/C

Answer 1

$$0.5mv^2=eU\\v^2=2(1.7588\cdot10^{11})(250)\\v=9.38\cdot10^6\frac{m}{s}$$