The capacitance of a parallel plate capacitor with air as a 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:\(\left(\varepsilon_0=8.85 \times 10^{-12} ~\text{C}^2 \text{N}^{-1} \text{m}^{-2}\right )\)

A parallel plate capacitor with cross-sectional area \(A\) and separation \(d\) has air between the plates. An insulating slab of the same area but the thickness of \(\frac{d}{2}\) is inserted between the plates as shown in the figure having a dielectric constant, \(K=4\). The ratio of new capacitance to its original capacitance will be:

A parallel-plate capacitor of area \(A\), plate separation \(d\) and capacitance \(C\) is filled with four dielectric materials having dielectric constants \(k_1, k_2,k_3\) and \(k_4\) as shown in the figure below. If a single dielectric material is to be used to have the same capacitance \(C\) in this capacitor, then its dielectric constant \(k\) is given by:
      
1. \( {k}={k}_1+{k}_2+{k}_3+3 {k}_4\)
2. \({k}=\frac{2}{3}\left({k}_1+{k}_2+{k}_3\right)+2 {k}_4\)
3. \({k}=\frac{2}{3} {k}_4\left(\frac{{k}_1}{{k}_1+{K}_4}+\frac{{k}_2}{{k}_2+{k}_4}+\frac{{k}_3}{{k}_3+{k}_4}\right)\)
4. \(\frac{1}{{k}}=\frac{1}{{k}_1}+\frac{1}{{k}_2}+\frac{1}{{k}_3}+\frac{3}{2 {k}_4}\)

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 the plate \(P\) is correctly shown by: 

   

1.		2.
3.		4.

Two parallel metal plates having charges +Q and –Q, face each other at a certain distance between them. If the plates are now dipped in the kerosene oil tank, the electric field between the plates will: