A dielectric slab of thickness 4 mm is placed between the plates of a parallel plate condenser. If the distance between plates is increased by 3.5 mm, the capacity of the condenser remains same. Find the dielectric of the medium.

8

Initial capacitance $=C_1=\varepsilon_0 A/d$

Final capacitance $=C_2=\frac{\varepsilon_0 A}{x+\left(d / \varepsilon_0\right)}$     (Here x = 3.5 mm)

As $C_1=C_2$, hence equating the values, we get

$\varepsilon_r=\frac{d}{d-x}=8$