Q.A metal plate is introduced between the plates of a charged parallel plate capacitor. What is its effect on the capacitance of the capacitor?

[!TLDR] A metal (conducting) slab is the k→∞ special case; inserting one of thickness t effectively shortens the gap to (d-t), so C …

A metal plate is a CONDUCTOR, which section 8.7.1 establishes always has zero field in its own interior -- exactly the limiting case $k\to\infty$ of the general dielectric-slab formula $C=\dfrac{\epsilon_0A}{(d-t)+t/k}$ from section 8.10.2. As $k\to\infty$, the term $t/k\to0$, leaving $C=\dfrac{\epsilon_0A}{d-t}=\dfrac{d}{d-t}C_0$ (special case 4 of that section) -- always GREATER than the bare $C_0=\epsilon_0A/d$, since a conducting slab of thickness t effectively shortens the working gap between the plates from d down …

Recognise a metal slab as the k→∞ limit of the general dielectric-slab f …

A frequent mistake is to treat inserting a conductor as somehow 'blocking' or reducing the capacitor's effect, rather than recognising it as effectively sh …