Q 12-02-028NEETJEE MainMedium
A parallel plate capacitor with plate separation $d$ has a metal slab of thickness $\dfrac{d}{2}$ inserted between its plates, parallel to them. The capacitance becomes
Answer: (A) $2$ times
The metal slab has no field inside it, so the effective gap is $d - \dfrac{d}{2} = \dfrac{d}{2}$. Capacitance doubles.
Solution by Sreeraj P, M.Sc Physics