Question

A parallel-plate capacitor is connected to a battery of electric potential difference V. If the plate separation is decreased, do the following quantities increase, decrease, or remain the same: (a) the capacitor’s capacitance, (b) the potential difference across the capacitor, (c) the charge on the capacitor, (d) the energy stored by the capacitor, (e) the magnitude of the electric field between the plates, and (f) the energy density of that electric field?

Short answer

1. The capacitor’s capacitance increases.
2. The potential difference across the capacitor is remains same.
3. The charge on the capacitor increases.
4. The energy stored by the capacitor increases.
5. The magnitude of electric field between the plates increases.
6. The energy density of that electric field increases.

Step-by-step solution

The given data

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

Understanding the concept of the capacitor properties

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, $C=\frac{{\in }_{0}A}{d}$ …(ii)

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

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

The electric field between the two plates,$E=\frac{1}{4\pi k{\in }_0}\frac{q}{r^2}or\frac{\delta }{k{\in }_0}$ …(iv)

where,is the charge density andis dielectric constant.

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

(a) Calculation of the capacitance of the capacitor

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.

(b) Calculation of the potential difference across the capacitor

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.

(c) Calculation of the charge on the capacitor

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.

(d) Calculation of the energy stored by the capacitor

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.

(e) Calculation of the magnitude of the electric field between the plates

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

Hence, the electric field increases.

(f) Calculation of the electric density of the field

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

Hence, the energy density also increases.