Answer:
The charge that will pass through the wire will be approx. 0.003C.
Explanation:
Two equations are relevant to answer this question:
(1) The charge (Q) over a capacitance (C), given a voltage (U):
[tex]Q = C\cdot U[/tex]
and
(2) the capacitance of a parallel-plate capacitor:
[tex]C=\varepsilon \frac{A}{d}[/tex]
with [tex]\varepsilon[/tex] being the dielectric constant, A the plate area and d their distance.
We are given the capacitance when air is present (80 micro-Farad, or uF). The dielectric constant for air/vacuum is 8.84 pico-Farad per meter. Then someone sneaky slides a slab of dielectric medium with dielectric constant of 20 pF/m and connects a source of 30V. By inserting the slab, they increased the capacitance by a factor of 20/8.84=2.26, making it 2.26*80=181uF. By connecting it to the source 30V, they charged our capacitor to a charge
[tex]Q = C\cdot V = 181\cdot 10^{-6}F\cdot 30V=5.43\cdot 10^{-3}C[/tex]
but, the sneaky person changes his mind and removes the slab, effectively changing C down to the original 80uF. There is now excess charge on the capacitor that wants to leave! The steady-state charge on the air-filled capacitor should be:
[tex]Q_0 = C_0\cdot V = 80\cdot 10^{-6}F\cdot 30V=2.4\cdot 10^{-3}C[/tex]
The charge that will pass through the wire will be the difference:
[tex]Q-Q_0=(5.43-2.40)\cdot 10^{-3}C\approx 3\cdot 10^{-3}C[/tex]
