Answer
Before:
$\Delta V_0 = 9.0$ Volts
$\vec{E}_0 = 9000$ N/C
$Q_0 = 20$ pC
After:
$\Delta V = 9.0$ Volts
$\vec{E} = 9000$ N/C
$Q = 62$ pC
Work Step by Step
Before the capacitor is filled with a dielectric, the values for potential difference, electric field, and charge can be calculated with the following equations: $\Delta V = \vec{E}d$ , $Q = C\Delta V$, and $C_0 = \displaystyle \frac{A\epsilon_0}{d}$
Part a.)
$\Delta V_0 = 9.0$ Volts (this is given to us)
$\vec{E}_0 = \displaystyle \frac{\Delta V}{d} = \frac{9}{0.0001} = 9000$ N/C (or Volts/m)
$Q_0 = C_0 \Delta V_0 = \displaystyle \frac{A\epsilon_0}{d}\Delta V_0 = \frac{(0.005)^2(8.85\cdot10^{-12})}{0.0001}(9) \approx 2.0\cdot 10^{-11}$ Coulombs = $20$ pC
Part b.)
Since the battery stays connected to the capacitor, the potential difference must stay the same.
$\Delta V = \Delta V_0 = 9.0$ Volts
Because potential difference remains constant, $\vec{E}$ also remains constant (Since everything in the equation stays constant in the equation$\Delta V = \vec{E}d$)
$\vec{E} = \vec{E}_0 = 9000$ N/C
The charge does change, however, but only because the capacitance must be multiplied by the dielectric constant (Table 26.1 gives the dielectric constant for Mylar as 3.1):
$Q = C\Delta V = \displaystyle \kappa\frac{A\epsilon_0}{d}\Delta V = (3.1)\frac{(0.005)^2(8.85\cdot10^{-12})}{0.0001}(9) \approx 6.2\cdot 10^{-11}$ Coulombs = $62$ pC
