Question

Calculate the work done by a 6.0-v battery as it charges a 7.6-μf capacitor in the flash unit of a camera.

Answer 1

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:
$W=U= \frac{1}{2}CV^2$
where 
$C=7.6 \mu F=7.6 \cdot 10^{-6} F$ is the capacitance of the capacitor
$V=6.0 V$ 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:
$W= \frac{1}{2}(7.6 \cdot 10^{-6} F)(6.0 V)^2=1.37 \cdot 10^{-4}J$