Answer:
5.3 nC
Explanation:
The initial charge stored on the capacitor is given by:
[tex]Q=CV[/tex] (1)
where
[tex]C=25 pF = 25\cdot 10^{-12}F[/tex] is the capacitance of the capacitor
[tex]V=100 V[/tex] is the potential difference across the capacitor
Substituting numbers into the equation, we have
[tex]Q=(25\cdot 10^{-12} F)(100 V)=25\cdot 10^{-10}C=2.5 nC[/tex]
When the Teflon slab is inserted between the plates, the capacitance of the capacitor is increased as follows:
[tex]C'=kC[/tex]
where k=2.1 is the dielectric constant of the Teflon. Since the voltage V remains constant, this means that the new charge stored by the capacitor (1) will be
[tex]Q'=C'V=kCV=kQ[/tex]
and so
[tex]Q'=(2.1)(2.5 nC)=5.3 nC[/tex]
