Answer
$1.34\times10^{-2}C$.
Work Step by Step
Calculate the energy using Equation 17–10, initial and final values.
$$PE_i=\frac{Q^2}{2C}$$
$$PE_f=\frac{(Q+\Delta Q)^2}{2C}$$
Find the energy difference.
$$PE_f-PE_i=\Delta PE=\frac{(Q+\Delta Q)^2}{2C}-\frac{Q^2}{2C}$$
Solve for the charge Q.
$$Q=\frac{C(\Delta PE)}{ \Delta Q}-\frac{1}{2}\Delta Q$$
$$Q=\frac{(17.0\times10^{-6}F)(15.2J)}{0.0130C}-\frac{1}{2}0.0130C =1.34\times10^{-2}C$$
