PHY capacitor 請答Q(e)and(f)

A parallel- plate capaitor has an area of 100cm^2 , a plate separation of 1.0cm and os chang initially to a potential of 100V (call this Vo )
The supply is disconnected and a slab of dielectric, 0.5cm thick and of relative permittivity 7 , is then placed between the plates as shown in Figure48.4

+Q
───────────

﹉﹉﹉﹉﹉﹉﹉
dielectic(0.5cm)
﹍﹍﹍﹍﹍﹍﹍

───────────
-Q

Figure48.4

calculate
(a) the capacitance Co before the slab is inserted,
(b) the charge on the plates Q ,
(c) the electric field strength in the gap between the plates and the dielectric,
(d) the electric field strength within the dielectic ,
(e) the potential difference between the plates after the dielectric is inserted,
(f) the capacitance when the sielectric has been inserted.


ans:
(a)Co=εoA/d=8.9*10^-12 / 10^-12=8.8*10^-12 F
(b)Q=CoVo=8.9*10^-12 * 100=8.9*10^-10 C
(c)Eo=Q/εoA=8.9*10^-10 / 8.9*10^-12 * 10^-2=10^4 Vm^-1
(d)E1=Q/εa = Eo/εr =10^4/7 (Eo/E1=εr)
(e)p.d. = 0.5*10^4+0.5*0.14*10^4 / 100 =57V
(f)C =εA/εr(d-t)+t = 7εoA/0.04 = 15.2*10^-12 F

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我不明白Q (e) and (f)
請解決我的問題呀
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