Answer
a) $\Delta Q=21.66C$
b) $\Delta Q=10.83C$
The amount of charges decreases by a factor of 2.
Work Step by Step
(a) We know that
$\Delta Q=\frac{W}{\epsilon}$
We plug in the known values to obtain:
$\Delta Q=\frac{260}{12}$
$\Delta Q=21.66C$
(b) We know that
$\Delta Q=\frac{W}{\epsilon}$
We plug in the known values to obtain:
$\Delta Q=\frac{160}{24}$
$\Delta Q=10.83C$
We can see that the amount of charges decreases by a factor of 2.
