solve if u can!!

Initially, assume the charge drawn by the capacitor to be Q. Next, let the area of the plates of the capacitors be A and the separation is given. So, the capacitance of the capacitor equals C=AEo/(5mm) (assume E0 to be the permittivity of free space). Let the potential difference applied across the capacitor be V. So, we get, Q=CV.----------------(1) Now, when we have inserted the mica sheet, we get the capacitor broken into three parts as follows: The first part- the part of the capacitor between the positive plate and the mica sheer. The second part-The mica sheet with dielectric constant K. The third part-The portion of the capacitor between the mica sheet and the negative plate of the capacitor. We can assume these parts to be individual capacitors placed in series. The first part has capacitance=AEo/(1.5mm)=the capacitance of the third part. The capacitance of the second part=A(KEo)/(2mm) (As permittivity changes to KEo inside a dielectric slab of dielectric constant K) By applying the formula for the series combination of capacitors, we get the final capacitance of the capacitor with the mica sheet inserted into it, as, A(KEo)/(2+3K) Even now the potential difference between the plates is equal to V but the charge drawn by the capacitor increases by 25% i.e. 1.25 times that of the previous one. so, [A(KEo)/(2+3K)] x V = 1.25Q Now by comparing this equation to the first equation i.e. Q=AEo/(5mm) We get the dielectric constant K=2 i hope this helps:)

When the $2mm$ mica sheet is added the capacitor draws $25\%$ more charges which means the new capacitance $C' = 1.25C$ .

Assuming the dielectric constant of air be $1$ , then we have $C=\frac{\epsilon_0A}{d}$ , where $A$ and $d=5mm$ are the area and separation of the plates respectively.

With the mica sheet added two capacitors with $3mm$ air and $2mm$ mica with dielectric constant $\epsilon_r$ as dielectric in series are formed. The combined capacitance $C'$ is given by:

$\cfrac{1}{C'}= \cfrac{4}{5} \times \cfrac{5}{\epsilon_0 A} = \cfrac{3}{\epsilon_0A}+\cfrac{2}{\epsilon_r \epsilon_0 A}$

$\Rightarrow 4 = 3 + \cfrac{2}{\epsilon_r}\quad \Rightarrow \epsilon_r = \boxed{2}$