In a parallel plate capacitor with air between the plates, each plate has an area of $\mathrm{}$\(6\times10^{-3}~\text{m}^2\), and the distance between the plates is \(3~\text{mm}\). The capacitance of the capacitor is:
1. \(16.12~\text{pF}\)
2. \(17.71~\text{pF}\)
3. \(15.01~\text{pF}\)
4. \(11.32~\text{pF}\)

The \(6\) \(\mu\)F capacitor is initially charged to \(2\) V (i.e. \(V_B-V_A=2\) V) while the \(3\) \(\mu\)F capacitor is uncharged. The switch is now closed. The final potential difference across the \(3\) \(\mu\)F capacitor will be:
    

1. \(4 \) V

2. \(\frac{4}{3} \) V

3. \(2\) V

4. \(\frac{8}{3} \) V

Three capacitors connected in series have a capacitance of \(9\) pF each. The potential difference across each capacitor if the combination is connected to a \(120\) V supply is:
1. \(10\) V
2. \(20\) V
3. \(30\) V
4. \(40\) V