The work done by the battery to charge the capacitor is equal to the electric potential energy stored in the capacitor at full charge, which is given by: [tex]W=U= \frac{1}{2}CV^2 [/tex] where [tex]C=7.6 \mu F=7.6 \cdot 10^{-6} F[/tex] is the capacitance of the capacitor [tex]V=6.0 V[/tex] is the voltage across the capacitor (which corresponds to the voltage of the battery)
By substituting the numbers into the equation, we find the work done: [tex]W= \frac{1}{2}(7.6 \cdot 10^{-6} F)(6.0 V)^2=1.37 \cdot 10^{-4}J [/tex]
