(A) 3.9
When a dielectric is inserted between the plates of a capacitor, the capacitance of the capacitor increases according to the equation:
[tex]C' = k C[/tex] (1)
where
C' is the final capacitance
k is the dielectric constant
C is the original capacitance
The capacitance is inversely proportional to the to voltage across the plates:
[tex]C=\frac{Q}{V}[/tex] (2a)
where Q is the charge stored and V the potential difference across the plates. We can rewrite C' (the capacitance of the capacitor filled with dielectric) as
[tex]C'=\frac{Q}{V'}[/tex] (2b)
Substituting (2a) and (2b) into (1), we find
[tex]V'=\frac{V}{k}[/tex] (3)
where
V = 45.0 V is the original voltage across the capacitor
V' = 11.5 V is the voltage across the capacitor filled with dielectric
Solving for k,
[tex]k=\frac{V}{V'}=\frac{45.0 V}{11.5 V}=3.9[/tex]
(B) 22.8 V
When the dielectric is partially pulled away, the system can be assimilated to a system of 2 capacitors in parallel, of which one of them is filled with dielectric and the other one is not.
Keeping in mind that the capacitance of a parallel-plate capacitor is proportional to the area of the plates:
[tex]C \propto A[/tex]
and in this case, the area of the capacitor filled with dielectric is just 1/3 of the total, we can write:
[tex]C_1 = \frac{2}{3}C\\C_2 = \frac{1}{3}kC[/tex]
where C1 is the capacitance of the part non-filled with dielectric, and C2 is the capacitance of the part filled with dielectric. The total capacitance of the system in parallel is
[tex]C'=C_1 + C_2 = \frac{2}{3}C+\frac{1}{3}kC=(\frac{2}{3}+\frac{1}{3}k)C[/tex]
Substituting,
[tex]C'=(\frac{2}{3}+\frac{1}{3}(3.9))C=1.97 C[/tex]
This is equivalent to a capacitor completely filled with a dielectric with dielectric constant k=1.97. Therefore, using again eq.(3), we find the new voltage:
[tex]V'=\frac{V}{k}=\frac{45.0 V}{1.97}=22.8 V[/tex]
