91Ó°ÊÓ

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

b) Potential difference, V=600 V

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

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,$C=\frac{{\mathrm{Î\mu }}_{0}A}{d}$ …(¾±)

The charge stored between the capacitor plates,$q=CV$ …(¾±¾±)

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

The electric field produced between the plates,$E=\frac{V}{d}$ …(¾±±¹)

The energy stored per unit volume in the field,$u=\frac{U}{V}$ …(±¹)

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

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

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

Hence, the value of the capacitance is 35.4 pF.

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

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

Hence, the value of the charge is 21 nC.

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

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

Hence, the value of the energy stored is $6.3\mathrm{μ´³}$.

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:

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

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

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:

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

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