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 $J/m^3$.

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 of $d=5.00mm$or $0.005m$and the potential difference is $V_{ab}=400.0\text{V}$.

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$\mu$by calculating the electric field between the two conductors.

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

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

Now,can be written as:

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

Calculating the energy density

Substituting the value of E in equation (i) we 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:

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

Therefore, the charge density$\mu$is$0.028{\text{J/m}}^{\text{3}}$