Answer
$2.96\times 10^{-15}J$
$1.85\times 10^4 eV$
Work Step by Step
The kinetic energy gained by the electron is the work done on it by the electric field.
The final potential is higher than the initial potential to accelerate the electron. It loses PE and gains KE. Use equation 17–2b.
$$V_{ba}=-\frac{W_{ba}}{q}$$
$$W_{ba}=-qV_{ba}=-(-1.60\times10^{-19}C)(1.85 \times 10^4V)=2.96\times 10^{-15}J$$
Multiply by $\frac{1eV}{1.60\times 10^{-19}J}$ to find the answer in electron volts, $1.85\times 10^4 eV$. As expected, this equals the potential difference in volts.
