The electrostatic force between the metal plates of an isolated parallel plate capacitor \(C\) having a charge \(Q\) and area \(A\) is:

A parallel plate air capacitor has capacitance \(C,\) the distance of separation between plates is \(d\) and potential difference \(V\) is applied between the plates. The force of attraction between the plates of the parallel plate air capacitor is:

The energy required to charge a parallel plate condenser of plate separation, \(d\) and plate area of cross-section, \(A\) such that the uniform electric field between the plates is \(E,\) is:
1. $\frac{1}{2}$ ${\mathrm{\epsilon }}_0{\mathrm{E}}^2/\mathrm{Ad}$

2. ${\mathrm{\epsilon }}_0{\mathrm{E}}^2/\mathrm{Ad}$

3. ${\mathrm{\epsilon }}_0{\mathrm{E}}^2\mathrm{Ad}$

4. $\frac{1}{2}$ ${\mathrm{\epsilon }}_0{\mathrm{E}}^2\mathrm{Ad}$

A parallel plate air capacitor is charged to a potential difference of V volts. After disconnecting the charging 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. does not change.

3. becomes zero.

4. increases.