(a) 3.33
For a capacitor with dielectric disconnected from the battery, the relationship between the voltage across the capacitor without the dielectric (V) and with the dielectric (V) is given by
[tex]V' = \frac{V}{k}[/tex]
where
k is the dielectric constant of the material
In this problem, we have
[tex]V' = 3.6 V[/tex]
[tex]V=12 V[/tex]
So we can re-arrange the formula to find the dielectric constant:
[tex]k=\frac{V}{V'}=\frac{12 V}{3.6 V}=3.33[/tex]
(b) The energy stored reduces by a factor 3.33
The energy stored in a capacitor is
[tex]U=\frac{1}{2}QV[/tex]
where
Q is the charge stored on the capacitor
V is the voltage across the capacitor
Here we can write the initial energy stored in the capacitor (without dielectric) as
[tex]U=\frac{1}{2}QV[/tex]
while after inserting the dielectric is
[tex]U'=\frac{1}{2}QV' = \frac{1}{2}Q\frac{V}{k}[/tex]
since Q, the charge, has not changed (the capacitor is disconnected, so the charge cannot flow away from the capacitor).
So the ratio between the two energies is
[tex]\frac{U'}{U}=\frac{\frac{1}{2}Q\frac{V}{k}}{\frac{1}{2}QV}=\frac{1}{k}[/tex]
which means
[tex]U' = \frac{U}{k}=\frac{U}{3.33}[/tex]
So, the energy stored has decreased by a factor 3.33.
(c) 5.5 V
Pulling the dielectric only partway so that it fills half of the space between the plates is equivalent to a system of 2 capacitors in parallel, each of them with area A/2 (where A is the original area of the plates of the capacitor), of which one of the two is filled with dielectric while the other one is not.
Calling
[tex]C=\frac{\epsilon_0 A}{d}[/tex] the initial capacitance of the capacitor without dielectric
The capacitance of the part of the capacitor of area A/2 without dielectric is
[tex]C_1 = \frac{\epsilon_0 \frac{A}{2}}{d}= \frac{C}{2}[/tex]
while the capacitance of the part of the capacitor with dielectric is
[tex]C_2 = \frac{k \epsilon_0 \frac{A}{2}}{d}= \frac{kC}{2}[/tex]
The two are in parallel, so their total capacitance is
[tex]C' = C_1 + C_2 = \frac{C}{2}+\frac{kC}{2}=(1+k)\frac{C}{2}=(1+3.33)\frac{C}{2}=2.17 C[/tex]
We also have that
[tex]V=\frac{Q}{C}=12 V[/tex] this is the initial voltage
So the final voltage will be
[tex]V' = \frac{Q}{C'}=\frac{Q}{2.17 C}=\frac{1}{2.17}V=\frac{12 V}{2.17}=5.5 V[/tex]
