What is the area of the plates of a \(2~\text{F}\) parallel plate capacitor, given that the separation between the plates is \(0.5~\text{cm}\)?
1. \(1100~\text{km}^2\)
2. \(1130~\text{km}^2\)
3. \(1110~\text{km}^2\)
4. \(1105~\text{km}^2\)

A parallel plate air capacitor is charged to potential difference \(V\). After disconnecting the battery, the distance between the plates of the capacitor is increased using an insulating handle. As a result the potential difference between the plates:
1. decreases.
2. increases.
3. becomes zero.
4. does not change.

In the given figure if \(V = 4~\text{volt}\) each plate of the capacitor has a surface area of \(10^{-2}~\text{m}^2\) and the plates are $\mathrm{}$\(0.1\times10^{-3}~\text{m}\) apart, then the number of excess electrons on the negative plate is:

           
1. \(5.15\times 10^{9}\)
2. \(2.21\times 10^{10}\)
3. \(3.33\times 10^{9}\)
4. \(2.21\times 10^{9}\)