Question

The parallel plate apparatus in Figure 6 accelerates electrons across a potential difference of 2963.6 V. Determine the speed with which the electron reaches the positively charged plate.  (b) How could this apparatus be modified to accelerate protons?  (c) If a proton was accelerated instead of an electron, with what speed would it reach the negative plate?

Answer 1

1)

$$mv^2=2eU\\v^2=2(1.7588\cdot10^{11})2963.6\\v=3.32\cdot10^7\frac{m}{s}$$

2) Change charged plates. 

3)

$$Mv^2=2eU\\v^2=2(9.5788\cdot10^{7})2963.6\\v=7.53\cdot10^5\frac{m}{s}$$