Problem 20.49

20.49.
When there is just air between the plates, the capacitance is given by
$C=\frac{Q}{V}$
$C=\frac{48 \mu C}{12 V}$
$C=4.00 \mu F$

b)
When the teflon is inserted, the charge on the plates isn't changed because the plates are isolated, so the capacitance and therefore the voltage changes.
$C = \kappa C_{0}$
$C = (2.1)(4 \mu F)$
$C = 8.40 \mu F$

c)
The charge was unaffected, so the voltage becomes
$V_{new} = \frac{Q}{C_{new}}$
$V_{new} = \frac{48 \mu C}{8.4 \mu F}$
V_new = 5.71 V