Question

A 5.80-µF, parallel-plate, air capacitor has a plate separation of 5.00 mm and is charged to a potential difference of 400 V. Calculate the energy density in the region between the plates, in units of J/${\mathit{m}}^{\mathbf{3}}$

Short answer

The energy density in the region between the plates is $0.028J/m^3$.

Step-by-step solution

Relation between the electric field and the separated conductors

Give,

It is given a capacitor C of 5.80 $\mu F$, and is separated by a distance ofor d = 5.00 mm and the potential difference is$V_{ab}=400.0\hspace{0.33em}V$.

Now, we know that the electric energy stored in a charged capacitor is equal to the amount of work required to charge it. Therefore, we can calculate the energy density uby calculating the electric field between the two conductors.

$u=\frac{1}{2}{\epsilon }_{\circ }{E}^2.............\left(i\right)$

Here, E is the electric field related to the separated distance between the two conductors.

Now, E can be written as:

$E=\frac{V}{d}.........\left(ii\right)$

Calculating the energy density:

Substituting the value of E in equationwe get:

$u=\frac{1}{2}{\epsilon }_{\circ }{\left(\frac{V}{d}\right)}^2$

Putting the values of$V,d,{\epsilon }_{\circ }$into the above equation we get:

$$\begin{gathered} u=\frac{1}{2}{\epsilon }_{\circ }{\left(\frac{V}{d}\right)}^2 \\ =\frac{1}{2}\left(8.854\times {10}^{-12}{C}^2/N.m^2\right){\left(\frac{400.0V}{0.005m}\right)}^2 \\ =0.028J/m^3 \end{gathered}$$

Therefore, the charge density u is $0.028J/m^3$.