Answer
$2.72\times 10^{-17}J$
$1.70\times 10^2 eV$
Work Step by Step
Calculate the work done by the electric field using equation 17–2b.
$$V_{ba}=-\frac{W_{ba}}{q}$$
$$W_{ba}=-qV_{ba}=-(1.60\times10^{-19}C)(-45V-125V)=2.72\times 10^{-17}J$$
Multiply by $\frac{1eV}{1.60\times 10^{-19}J}$ to find the answer in electron volts, $1.70\times 10^2 eV$.