A parallel plate capacitor of capacitance \(90~\text{pF}\) is connected to a battery of emf \(20~\text{V}\). If a dielectric material of dielectric constant \(K=\frac{5}{3}\) is inserted between the plates, the magnitude of the induced charge will be:
1. \(1.2~\text{nC}\)
2. \(0.3~\text{nC}\)
3. \(2.4~\text{nC}\)
4. \(0.9~\text{nC}\)

Voltage rating of a parallel plate capacitor is \(500~\text{V}\). Its dielectric can withstand a maximum electric field of \(10^6~\text{V/m}\). The plate area is \(10^{-4}~\text{m}^2\). What is the dielectric constant if the capacitance is \(15~\text{pF}\)? (given \(\varepsilon_0=8.86 \times 10^{-12}~ \text{C}^2 / \text{Nm}^2\))
1. \(6.2\)
2. \(3.8\)
3. \(4.5\)
4. \(8.5\)

The parallel combination of two air filled parallel plate capacitors of capacitance \(C\) and \(nC\) is connected to a battery of voltage, \(V\). When the capacitors are fully charged, the battery is removed and after that a dielectric material of dielectric constant \(K\) is placed between the two plates of the first capacitors. The new potential difference of the combined system is: 
1. \( \frac{V}{K+n} \)
2. \(\frac{n V}{K+n} \)
3. \(\frac{(n+1) V}{(K+n)} \)
4. \({V}\)

Two identical parallel plate capacitors, of capacitance \(C\) each, have plates of area \(A\), separated by a distance \(d\). The space between the plates of the two capacitors is filled with three dielectrics, of equal thickness and dielectric constants \(K_1,K_2\) and \(K_3\). The first capacitor is filled as shown in Fig \(\mathrm{I}\), and the second one is filled as shown in Fig \(\mathrm{II}\).
If these two modified capacitors are charged by the same potential \(V\), the ratio of the energy stored in the two, would be (\(E_1\) refers to capacitor (\(\mathrm{I}\)) and \(E_2\) to capacitor (\(\mathrm{II}\))):

                         
1. \(\frac{E_1}{E_2}=\frac{(K_1+K_2+K_3)(K_3K_1+K_2K_3+K_2K_1)}{K_1K_2K_3}\)

2. \(\frac{E_1}{E_2}=\frac{K_1K_2K_3}{(K_1+K_2+K_3)(K_3K_1+K_2K_3+K_2K_1)}\)

3. \(\frac{E_1}{E_2}=\frac{(K_1+K_2+K_3)(K_3K_1+K_2K_3+K_2K_1)}{9K_1K_2K_3}\)

4. \(\frac{E_1}{E_2}=\frac{9K_1K_2K_3}{(K_1+K_2+K_3)(K_3K_1+K_2K_3+K_2K_1)}\)