Answer:
V = 9.33 V
Explanation:
By definition, the capacitance of a capacitor, is the proportion between the charge on one of charged surfaces and the potential difference between these surfaces, as follows:
[tex]C = \frac{Q}{V}[/tex]
for a parallel-plate type capacitor, it can be showed, applying Gauss'Law to the a gaussian surface with the shape of a pillbox parallel to the surface of one of the plates, half outside the surface, half inside it, that the capacitance can be expressed as follows:
[tex]C =\frac{\epsilon*A}{d}[/tex]
As we can see, if a slab of paraffin, with dielectric constant of 2.25, is inserted, the capacitance will be increased in the same factor:
[tex]Cf = 2.25 * Co[/tex]
Now, if the capacitor, once charged, is disconnected from the battery, the charge Q will remain constant.
So, if the capacitance will be increased 2.25 times, the only way to do this is that the voltage between plates be reduced in the same factor:
[tex]Cf = 2.25 * Co = 2.25* \frac{Q}{Vo} = \frac{Q}{Vf} \\ Vf= \frac{Vo}{2.25} =\frac{21.0V}{2.25} = 9.33 V[/tex]
