Answer:
0.275 nF
Explanation:
First of all, let's calculate the initial charge stored by the capactiro when it is in air. This is given by:
[tex]Q_0 = C_0 V[/tex]
where
[tex]C_0 = 25 pF=25\cdot 10^{-12} F[/tex] is the initial capacitance of the capacitor
[tex]V=100 V[/tex] is the voltage of the battery
Substituting, we find
[tex]Q_0 = (25\cdot 10^{-12}F)(100 V)=25\cdot 10^{-10} F=0.25 nF[/tex]
When the Teflon slab is inserted between the plates, the capacitance changes according to:
[tex]C'=kC_0[/tex]
where
[tex]k=2.1[/tex] is the dielectric constant of Teflon. Therefore, the new charge stored on the capacitor will be:
[tex]Q'=C' V=(kC_0)V=(2.1)(25\cdot 10^{-12} F)(100 V)=0.525 nF[/tex]
And so, the change in the charge on the capacitor is:
[tex]\Delta Q=Q'-Q=0.525 nF-0.25 nF=0.275 nF[/tex]
