Q 1 :    

The capacitance of a parallel plate capacitor with air as medium is $6$ $\mu \text{F}$. With the introduction of a dielectric medium, the capacitance becomes $30$$\mu \text{F}$. The permittivity of the medium is (${\epsilon }_0=8.85\times {10}^{-12} {\text{C}}^2{\text{N}}^{-1}{\text{m}}^{-2}$)            [2020]
 

• $0.44\times {10}^{-13} {\text{C}}^2{\text{N}}^{-1}{\text{m}}^{-2}$

   

• $1.77\times {10}^{-12} {\text{C}}^2{\text{N}}^{-1}{\text{m}}^{-2}$

   

• $0.44\times {10}^{-10} {\text{C}}^2{\text{N}}^{-1}{\text{m}}^{-2}$

   

• $5.00 {\text{C}}^2{\text{N}}^{-1}{\text{m}}^{-2}$

   

Q 2 :    

Two thin dielectric slabs of dielectric constants $K_1$ and $K_2$$(K_1<K_2)$ are inserted between plates of a parallel plate capacitor, as shown in the figure. The variation of electric field $E$ between the plates with distance $d$ as measured from plate $P$ is correctly shown by         [2014]