Answer
Plate B is at an electric potential 4030 volts higher than that of plate A.
Work Step by Step
The kinetic energy gained by the electron is the work done on it by the electric field.
The final potential, at plate B, is higher than the initial potential (plate A) to accelerate the electron. It loses PE and gains KE. Use equation 17–2b.
$$V_{ba}=-\frac{W_{ba}}{q}$$
$$V_{ba}=-\frac{6.45\times10^{-16}J}{-1.60\times10^{-19}C}=4030V$$
