veksetz

2022-01-04

A capacitor consists of two 6.0-cm-diameter circular plates separated by 1.0 mm. The plates are charged to 150 V, then the battery is removed.
a. How much energy is stored in the capacitor?
b. How much work must be done to pull the plates apart to where the distance between them is 2.0 mm?

ramirezhereva

Expert

We are given:
$D=6cm$, $d=1mm$, $V=150V$
a) The energy stored in the capacitor will be obtained using this formula:
$U=\frac{1}{2}C{V}^{2}$
Where $C$ is the capacitance of the capacitor and $D$ is the potential
The capacitance we can get from the following formula:
$C=\frac{{ϵ}_{0}A}{d}$
A capacitor consists of circular plates, so its surface is:
$A=\pi ×{r}^{2}$
$r=\frac{D}{2}=\frac{6}{2}=3cm$
${ϵ}_{0}$ is the dialectric constznt of vacuum and has the following value:
${ϵ}_{0}=8.854×{10}^{-12}\frac{F}{m}$
Thus, the first formula becomes:
$U=\frac{1}{2}×\frac{{ϵ}_{0}\pi {r}^{2}}{d}×{V}^{2}$
Now, convert cm and mm into m
Finally, we have:
$U=\frac{1}{2}×\frac{8.854×{10}^{-12}\pi ×{\left(3×{10}^{-2}\right)}^{2}}{1×{10}^{-3}}×{\left(150\right)}^{2}$
$U=0.28×{10}^{-6}J$
b)Here the distance between the plates increases to ${d}_{1}=2mm$
The work required to achieve will be obtained as the energy difference:
$W=U-{U}_{1}$
${U}_{1}=\frac{1}{2}×\frac{8.854×{10}^{-12}\pi ×{\left(3×{10}^{-2}\right)}^{2}}{2×{10}^{-3}}×{150}^{2}=0.14×{10}^{-6}J$
And we have:
$W=0.28×{10}^{-6}-0.14×{10}^{-6}$
$W=0.14×{10}^{-6}J$

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