91影视

Give,

It is given a capacitor C of $5.80渭F$, 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$渭$by calculating the electric field between the two conductors.

$u=\frac{1}{2}{蔚}_{鈭�}{E}^2.............\left(i\right)$

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

Now,can be written as:

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

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

$u=\frac{1}{2}{蔚}_{鈭�}{\left(\frac{V}{d}\right)}^2$

Putting the values of $V,d,{蔚}_{鈭�}$into the above equation we get:

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

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