91Ó°ÊÓ

1. The potential difference of the battery is V.
2. The plate separation is decreases.

Using Eq. 25-9, 25-1, 25-22, 25-28, and 25-25, we can find the capacitor’s capacitance and the potential difference across the capacitor, the charge on the capacitor, the energy stored by the capacitor, the magnitude of the electric field between the plates, the energy density of that electric field increases, decreases or same with the decrease in separation respectively.

Formulae:

The charge within the plates of the capacitor, q=CV …(i)

The capacitance between the plates with air in the medium, $C=\frac{{∈}_{0}A}{d}$ …(¾±¾±)

where,${∈}_0$is permittivity constant, A is the plate area and d is the separation between the plates.

The potential energy of the capacitor,$U=\frac{1}{2}C{V}^2$ …(¾±¾±¾±)

The electric field between the two plates,$E=\frac{1}{4\mathrm{Ï€}k{∈}_0}\frac{q}{r^2}or\frac{\mathrm{δ}}{k{∈}_0}$ …(¾±±¹)

where,is the charge density andis dielectric constant.

The electric density due to the electric field, $u=\frac{1}{2}{∈}_0{E}^2$ …(±¹)

The capacitance of the parallel-plate capacitors can be given using equation (ii).

From this equation, we can conclude that if d decreases, the capacitor’s capacitance increases.

Since, the capacitor is still connected to the battery and the plate separation doesn’t affect the potential difference will not change.

Therefore, potential difference across the capacitor remains the same.

From equation (i) of charge, we can conclude that if capacitance C increases and the potential difference remains same, then the charge should get increased.

Hence, the charge on the capacitor increases with increasing plate separation.

From equation (iii) of potential energy, we can interpret that if capacitance C gets increased and the potential difference remains the same.

Hence, the energy of the capacitor increases.

From equation (iv), we can interpret that if charge q gets increased the charge density$\mathrm{δ}$is also increased.

Hence, the electric field increases.

From equation (v) of the energy density, we can infer that if electric field is increased.

Hence, the energy density also increases.