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 $\mathrm{V}=4 v\mathrm{olt}$, each plate of the capacitor has a surface area of ${10}^{-\mathrm{2 }}{\mathrm{m}}^2$ and the plates are $0.1\times {10}^{-\mathrm{3 }}\mathrm{m}$ 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. $2.21\times {10}^9$