## Fundamentals of Physics Extended (10th Edition)

$W=2.30\times 10^{-30}J$
The initial electric energy is given as $U_i=\frac{Kq^2}{a}$ The final electric energy is given as $U_f=3\times \frac{Kq^2}{a}$ We know that $W=\Delta U=U_f-U_i$ We plug in the known values to obtain: $W=3\times \frac{Kq^2}{a}-\frac{Kq^2}{a}=2\times \frac{Kq^2}{a}$ We plug in the known values to obtain: $W=2\times \frac{9\times 10^9(1.6\times 10^{-19})^2}{2\times 10^{-6}}=2.30\times 10^{-30}J$