Question

An isolated parallel plate capacitor is charged upto a certain potential difference.When a 3mm thick slab is introduced between the plates then in order to maintain the same potential difference, the distance between the plates is increased by 2.4mm. Find the dielectric constant of the slab. (assume charge remains constant)

Answer 1

Before inserting dielectric slab the capacitance $C_i=\frac{A{ϵ}_0}{d}$ where $A=$ area of plate and $d=$ separation between plates.
After inserting dielectric slab the capacitance $C_f=\frac{A{ϵ}_0}{d^{\prime }-t(1-\frac{1}{K})}$ where $K=$ dielectric constant and $t=$ thickness of slab.
As maintain same potential and isolated capacitor so the capacitance will be same for both case. i.e
$C_i=C_f$
$\frac{A{ϵ}_0}{d}=\frac{A{ϵ}_0}{d^{\prime }-t(1-\frac{1}{K})}\Rightarrow d=d^{\prime }-t(1-\frac{1}{K})$

Here, $d^{\prime }=(d+2.4)mm$ and $t=3mm$

Thus, $d=(d+2.4)-3(1-\frac{1}{K})$

$\Rightarrow 1-\frac{1}{K}=\frac{2.4}{3}=0.8$

$\Rightarrow \frac{1}{K}=1-0.8=0.2\Rightarrow K=5$