Question

A parallel-plate air-filled capacitor having area 44 cm2 and plate spacing 5.0 mm is charged to a potential difference of 850 V. Find the following values.
(a) the capacitance (pF)

(b) the magnitude of the charge on each plate (nC)

(c) the stored energy (μJ)

(d) the electric field between the plates (V/m)

(e) the energy density between the plates (J/m³)

Answer 1

Concepts and reason

The required concepts to solve the problem are the capacitance of the capacitor, charge on the plates of the capacitor, the energy stored in the capacitor, the electric field between the plates of the capacitor and the energy density between the plates of the capacitor.

First, using the relation between the plate separation, area of cross-section and the capacitance, find the capacitance of the capacitor. Then, the charge on the plate of the capacitor is found by taking the product of the capacitance and the potential difference.

Then, using the relation between the capacitance, the energy stored in the capacitor and the potential difference, the energy stored in the capacitor can be found. The ratio of the potential difference and the plate separation can give the electric field between the plates. Then, from the electric field, the energy density of the capacitor can be found.

Fundamentals

A parallel plate capacitor is a capacitor formed by two metal plates separated by a distance. The capacitance of a parallel plate capacitor is,

Here, is the permittivity of free space, is the area of cross-section and is the separation distance between the plates.

A capacitor is used to store charges. The charge on each plate of the capacitor is,

$Q=CV $

Here, is the potential difference between the plates of the capacitor and Q is the charge.

The energy stored in a capacitor is found from the capacitance and the potential difference.

Here, U is the energy.

The electric field between the plates is,

Here, E is the electric field.

The energy density of the capacitor is,

Here, is the energy density.

(a)

The capacitance of a parallel plate capacitor is,

Substitute for , for and for .

(b)

The charge on each plate of the capacitor is,

$Q=CV $

Substitute for and for .

(c)

The energy stored in the capacitor is,

Substitute for and for .

(d)

The electric field between the plates of the capacitor is,

Substitute for and for .

(e)

The energy density between the plates of the capacitor is,

Substitute for and for.

Ans: Part a

The capacitance of a parallel plate capacitor is.

Part b

The charge on each plate of the capacitor is .

Part c

The energy stored in the capacitor is.

Part d

The electric field between the plates of the capacitor is .

Part e

The energy density between the plates of the capacitor is .