Question

A parallel-plate air-filled capacitor having area$\mathbf{40}{\mathbf{cm}}^{\mathbf{2}}$ and plate spacing 1.0 mmis charged to a potential difference of 600 V(a) Find the capacitance,

(b) Find the magnitude of the charge on each plate, (c) Find the stored energy,

(d) Find the electric field between the plates, and (e) Find the energy density between the plates.

Short answer

a) The value of capacitance is 35.4 pF

b) The magnitude of charge on each plate is 21 nC

c) The energy stored in capacitor is$6.3\mathrm{\mu J}$

d) The electric field between the plates is $6.0\times {10}^5\mathrm{V}/\mathrm{m}$

e) The energy density between the plates is$1.57\mathrm{J}/{\mathrm{m}}^3$

Step-by-step solution

The given data

a) Area$A=40c{m}^{2}or4.0\times {10}^{-3}{m}^2$

b) Potential difference, V=600 V

c) Distance between plates,$\mathrm{d}=1.0\mathrm{mm}\mathrm{or}{10}^{-3}\mathrm{m}$

Understanding the concept of the capacitance and energy

We use the formula of the capacitance of the parallel plate capacitor to find capacitance. Using the relation between charge and capacitance, we can find the magnitude of the charge. Using the formula of energy stored in the capacitor, we can find the energy stored in the capacitor. We use the relation of the electric field, the distance, and the potential difference to find the magnitude of the electric field. Using the energy density formula, we can find density.

Formulae:

The capacitance between the two plates,$C=\frac{{\epsilon }_{0}A}{d}$ …(i)

The charge stored between the capacitor plates,$q=CV$ …(ii)

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

The electric field produced between the plates,$E=\frac{V}{d}$ …(iv)

The energy stored per unit volume in the field,$u=\frac{U}{V}$ …(v)

(a) Calculation of the capacitance value

A parallel plate capacitor with flat parallel plates of areaand spacinghas capacitance that can be calculated using equation (i) as follows:

(${\mathrm{\epsilon }}_0=8.85\times {10}^{-12}\frac{{\mathrm{C}}^2}{\mathrm{N}.{\mathrm{m}}^2}$is permittivity of free space)

$$\begin{gathered} \mathrm{C}=\frac{\left(8.85\times {10}^{-12}{\mathrm{C}}^2/\mathrm{N}.{\mathrm{m}}^2\right)\left(4.0\times {10}^{-3}{\mathrm{m}}^2\right)}{{10}^{-3}\mathrm{m}} \\ =3.54\times {10}^{-11}\mathrm{F} \\ =35.4\mathrm{pF} \end{gathered}$$

Hence, the value of the capacitance is 35.4 pF.

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

Using the given data in equation (ii), we can get the magnitude of the charge as follows:

$$\begin{gathered} q=\left(35.4pF\right)\left(600V\right) \\ =2.1\times {10}^{-8}C \\ =21nC \end{gathered}$$

Hence, the value of the charge is 21 nC.

(c) Calculation of the energy stored in capacitor

The amount of energy stored in the capacitor can be given using the given data in equation (iii) as follows:

$$\begin{gathered} U=\frac{1}{2}\left(35.4pF\right){\left(600V\right)}^2 \\ =\frac{1}{2}\times 35.4\times {10}^{-12}F\times {\left(600V\right)}^2 \\ =6.3\times {10}^{-6}J \\ =6.3\mu J \end{gathered}$$

Hence, the value of the energy stored is $6.3\mathrm{\mu J}$.

(d) Calculation of the electric field between the plates

In order to accelerate the charge in electric field from negative plate to positive plate, we apply the potential difference between the plates and the value of the electric field can be given using equation (iv) as follows:

$$\begin{gathered} E=\frac{600V}{{10}^{-3}m} \\ =6.0\times {10}^5\mathrm{V}/\mathrm{m} \end{gathered}$$

Hence, the value of the electric field is role="math" localid="1661341070430" $6.0\times {10}^5\mathrm{V}/\mathrm{m}$.

(e) Calculation of the energy density between the plates

It is energy per unit volume.

The volume of parallel plate using its area and distance can be given by:

V=Ad

Thus, the value of the energy density can be given using the above volume value in equation (v) as follows:

$$\begin{gathered} u=\frac{6.3\times {10}^{-6}\mathrm{J}}{\left(4.0\times {10}^{-3}{\mathrm{m}}^2\right)\left({10}^{-3}\mathrm{m}\right)} \\ =1.57\mathrm{J}/{\mathrm{m}}^3 \end{gathered}$$

Hence, the value of the energy density is $1.57\mathrm{J}/{\mathrm{m}}^3$.