Susan Nall

2022-01-05

A $5.80-\mu F$, 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 $\frac{J}{{m}^{3}}$.

Janet Young

Required
We are asked to find the energy density u in $\frac{J}{{m}^{3}}$
Solution:
The electric potential energy stored in a charged capacitor is just equal to the amount to work required to charge it that is, to separate opposite charges and place them on different conductors. We can calculate the energy density u by calculating the electric field between the two conductors as in equation 24.11 $u=\frac{1}{2}{ϵ}_{\circ }{E}^{2}$(1)
Where the electric field E is related to the separated distance between the two conductors as next $E=\frac{V}{d}$
If we plug the expression of E into equation (1) we would get the next relation between the energy density, potential difference and the separated distance as next $u=\frac{1}{2}{ϵ}_{\circ }{\left(\frac{V}{d}\right)}^{2}$(2)
Now we can plug our values for V, d and ${ϵ}_{\circ }$ into equation (2) to get u
$u=\frac{1}{2}{ϵ}_{\circ }{\left(\frac{V}{d}\right)}^{2}$
$=\frac{1}{2}\left(8.854×{10}^{-12}\frac{{C}^{2}}{N}\cdot {m}^{2}\right){\left(\frac{400.0V}{0.005m}\right)}^{2}$
$=0.028\frac{J}{{m}^{3}}$
As shown by equation (1) we can think of the energy as being stored in the field in the region between the plates.

Navreaiw

The expression of electrostatic energy density is ${u}_{E}=\frac{1}{2}{ϵ}_{0}{E}^{2}$
Here, $E=\frac{V}{d}=\frac{400}{5×{10}^{-3}}=8×{10}^{4}\frac{V}{m}$
Thus, ${u}_{E}=0.5×\left(8.85×{10}^{-12}\right)×{\left(8×{10}^{4}\right)}^{2}=2.83×{10}^{-2}\frac{J}{{m}^{3}}$

karton

The expression of electrostatic energy density is ${u}_{E}=\frac{1}{2}{ϵ}_{0}{E}^{2}$
Here, $E=V/d=\frac{400}{5×{10}^{-3}}=8×{10}^{4}V/m$
Thus, ${u}_{E}=0.5×\left(8.85×{10}^{-12}\right)×\left(8×{10}^{4}{\right)}^{2}=2.83×{10}^{-2}J/{m}^{3}$

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