Problem 20.63

20.63. a)
The system may be thought of as two parallel plate capacitors connected in series. In that case, the capacitance of each one is given by
$C_{1} = \frac{\epsilon_{0}A}{t_{1}}$
$C_{2} = \frac{\epsilon_{0}A}{t_{2}}$
Where t₁ and t₂ are the separations between the plates and the conducting slab. Using the formula for capacitors in series
$\frac{1}{C_{eff}} = \sum_{i} \frac{1}{C_{i}}$
we find that the effective capacitence of the system is given by
$\frac{1}{C_{eff}} = \frac{t_{1}+t_{2}}{\epsilon_{0}A}$
$C_{eff} = \frac{\epsilon_{0}A}{t_{1} + t_{2}}$
The quantity t₁ + t₂ is the space between capacitor plates that is not filled with the conductor. This is also given as s-d.
$C_{eff} = \frac{\epsilon_{0}A}{s-d}$