veksetz

## Answered question

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?

### Answer & Explanation

ramirezhereva

Beginner2022-01-05Added 28 answers

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:
$A=\pi ×{r}^{2}$
And radius is:
$r=\frac{D}{2}=\frac{6}{2}=3cm$
${ϵ}_{0}$ 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:
$W=0.28×{10}^{-6}-0.14×{10}^{-6}$
$W=0.14×{10}^{-6}J$

Nick Camelot

Skilled2023-06-19Added 164 answers

a. The energy stored in a capacitor can be calculated using the formula:
$E=\frac{1}{2}C{V}^{2}$
where $E$ is the energy stored, $C$ is the capacitance, and $V$ is the voltage. The capacitance of a parallel plate capacitor is given by:
$C=\frac{{ϵ}_{0}A}{d}$
where ${ϵ}_{0}$ is the permittivity of free space, $A$ is the area of the plates, and $d$ is the separation distance between the plates. Substituting these values into the formulas:
$C=\frac{{ϵ}_{0}\pi {r}^{2}}{d}$
$E=\frac{1}{2}\frac{{ϵ}_{0}\pi {r}^{2}}{d}{V}^{2}$
where $r$ is the radius of the plates.
Substituting the given values:
$E=\frac{1}{2}\frac{\left(8.85×{10}^{-12}\phantom{\rule{0.167em}{0ex}}\text{F/m}\right)\left(\pi \left(0.03\phantom{\rule{0.167em}{0ex}}\text{m}{\right)}^{2}\right)}{0.001\phantom{\rule{0.167em}{0ex}}\text{m}}\left(150\phantom{\rule{0.167em}{0ex}}\text{V}{\right)}^{2}$
b. The work done to pull the plates apart can be calculated using the formula:
$W=\frac{1}{2}C{V}_{1}^{2}-\frac{1}{2}C{V}_{2}^{2}$
where $W$ is the work done, $C$ is the capacitance, ${V}_{1}$ is the initial voltage, and ${V}_{2}$ is the final voltage. Substituting the given values:
$W=\frac{1}{2}\left(\frac{{ϵ}_{0}\pi {r}^{2}}{d}\left(150\phantom{\rule{0.167em}{0ex}}\text{V}{\right)}^{2}\right)-\frac{1}{2}\left(\frac{{ϵ}_{0}\pi {r}^{2}}{d}\left(150\phantom{\rule{0.167em}{0ex}}\text{V}{\right)}^{2}\right)$
$W=0$
Therefore, no work is required to pull the plates apart, as the energy stored in the capacitor remains the same.

Eliza Beth13

Skilled2023-06-19Added 130 answers

Step 1: a. We can use the following formula to determine how much energy is held in a capacitor:
$U=\frac{1}{2}C{V}^{2}$ where $U$ is the energy stored, $C$ is the capacitance, and $V$ is the voltage.
The capacitance $C$ of a parallel plate capacitor is given by:
$C=\frac{{\epsilon }_{0}A}{d}$ where ${\epsilon }_{0}$ is the vacuum permittivity, $A$ is the area of the plates, and $d$ is the distance between the plates.
Given that the plates have a diameter of 6.0 cm, we can find the area $A$ using the formula:
$A=\pi {r}^{2}$ where $r$ is the radius of the plates.
Substituting these values into the equations, we have:
$A=\pi {\left(\frac{6.0\phantom{\rule{0.167em}{0ex}}\text{cm}}{2}\right)}^{2}$
$d=1.0\phantom{\rule{0.167em}{0ex}}\text{mm}$
$V=150\phantom{\rule{0.167em}{0ex}}\text{V}$
Calculating $A$, $C$, and $U$:
$A=\pi {\left(\frac{6.0\phantom{\rule{0.167em}{0ex}}\text{cm}}{2}\right)}^{2}=\pi {\left(\frac{0.06\phantom{\rule{0.167em}{0ex}}\text{m}}{2}\right)}^{2}$
$d=1.0\phantom{\rule{0.167em}{0ex}}\text{mm}=0.001\phantom{\rule{0.167em}{0ex}}\text{m}$
$C=\frac{{\epsilon }_{0}A}{d}=\frac{8.85×{10}^{-12}\phantom{\rule{0.167em}{0ex}}\text{F/m}×\pi {\left(\frac{0.06\phantom{\rule{0.167em}{0ex}}\text{m}}{2}\right)}^{2}}{0.001\phantom{\rule{0.167em}{0ex}}\text{m}}$
$U=\frac{1}{2}C{V}^{2}=\frac{1}{2}\left(\frac{8.85×{10}^{-12}\phantom{\rule{0.167em}{0ex}}\text{F/m}×\pi {\left(\frac{0.06\phantom{\rule{0.167em}{0ex}}\text{m}}{2}\right)}^{2}}{0.001\phantom{\rule{0.167em}{0ex}}\text{m}}\right)×\left(150\phantom{\rule{0.167em}{0ex}}\text{V}{\right)}^{2}$
Now we can calculate the values.
Step 2: b. To find the work done in pulling the plates apart, we need to consider the change in potential energy.
The work done is given by:
$W={U}_{\text{final}}-{U}_{\text{initial}}$
Given that the initial distance between the plates is 1.0 mm and the final distance is 2.0 mm, we can calculate the initial and final energies using the same formula as in part a.
Substituting the values into the equation:
$W={U}_{\text{final}}-{U}_{\text{initial}}=\left(\frac{1}{2}C{V}_{\text{final}}^{2}\right)-\left(\frac{1}{2}C{V}_{\text{initial}}^{2}\right)$ where ${V}_{\text{final}}$ is the voltage when the plates are pulled apart to a distance of 2.0 mm and ${V}_{\text{initial}}$ is the initial voltage.
Now we can calculate the values.

Mr Solver

Skilled2023-06-19Added 147 answers

Answer:
$W=\frac{1}{2}\left(\frac{{\epsilon }_{0}\pi {\left(\frac{6.0}{2}\right)}^{2}}{1.0}\right){\left(\frac{{Q}_{i}}{\frac{{\epsilon }_{0}\pi {\left(\frac{6.0}{2}\right)}^{2}}{2.0}}-\frac{{Q}_{i}}{\frac{{\epsilon }_{0}\pi {\left(\frac{6.0}{2}\right)}^{2}}{1.0}}\right)}^{2}\phantom{\rule{0.167em}{0ex}}\text{J}$
Explanation:
a. To calculate the energy stored in the capacitor, we can use the formula for the energy stored in a capacitor:
$E=\frac{1}{2}C{V}^{2}$
where $E$ is the energy stored, $C$ is the capacitance, and $V$ is the voltage.
First, let's calculate the capacitance of the capacitor. The capacitance of a parallel plate capacitor can be given as:
$C=\frac{{\epsilon }_{0}A}{d}$
where ${\epsilon }_{0}$ is the vacuum permittivity, $A$ is the area of the plates, and $d$ is the separation distance between the plates.
Given that the diameter of each circular plate is 6.0 cm, we can find the area of each plate using the formula for the area of a circle:
$A=\pi {r}^{2}$
where $r$ is the radius of the plate.
Substituting the values, we get:
$A=\pi {\left(\frac{6.0\phantom{\rule{0.167em}{0ex}}\text{cm}}{2}\right)}^{2}$
Now, let's calculate the capacitance:
$C=\frac{{\epsilon }_{0}\pi {\left(\frac{6.0\phantom{\rule{0.167em}{0ex}}\text{cm}}{2}\right)}^{2}}{1.0\phantom{\rule{0.167em}{0ex}}\text{mm}}$
Now, we can substitute the given values of capacitance and voltage into the energy formula:
$E=\frac{1}{2}\left(\frac{{\epsilon }_{0}\pi {\left(\frac{6.0\phantom{\rule{0.167em}{0ex}}\text{cm}}{2}\right)}^{2}}{1.0\phantom{\rule{0.167em}{0ex}}\text{mm}}\right)\left(150\phantom{\rule{0.167em}{0ex}}\text{V}{\right)}^{2}$
Simplifying the expression, we have:
$E=\frac{1}{2}\left(\frac{{\epsilon }_{0}\pi {\left(\frac{6.0}{2}\right)}^{2}}{1.0}\right)\left(150{\right)}^{2}\phantom{\rule{0.167em}{0ex}}\text{J}$
b. To calculate the work done in pulling the plates apart, we can use the formula for the work done on a capacitor:
$W=\frac{1}{2}C{\left(\frac{{Q}_{f}}{{C}_{f}}-\frac{{Q}_{i}}{{C}_{i}}\right)}^{2}$
where $W$ is the work done, ${Q}_{f}$ is the final charge on the capacitor, ${C}_{f}$ is the final capacitance, ${Q}_{i}$ is the initial charge on the capacitor, and ${C}_{i}$ is the initial capacitance.
Since the battery is removed, the charge on the capacitor remains constant. Therefore, ${Q}_{f}={Q}_{i}$.
Let's denote the initial separation distance as ${d}_{i}$ and the final separation distance as ${d}_{f}$.
Using the formula for capacitance, we can express the initial and final capacitance as:
${C}_{i}=\frac{{\epsilon }_{0}A}{{d}_{i}}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}{C}_{f}=\frac{{\epsilon }_{0}A}{{d}_{f}}$
Substituting the given values, we get:
${C}_{i}=\frac{{\epsilon }_{0}\pi {\left(\frac{6.0\phantom{\rule{0.167em}{0ex}}\text{cm}}{2}\right)}^{2}}{1.0\phantom{\rule{0.167em}{0ex}}\text{mm}}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}{C}_{f}=\frac{{\epsilon }_{0}\pi {\left(\frac{6.0\phantom{\rule{0.167em}{0ex}}\text{cm}}{2}\right)}^{2}}{2.0\phantom{\rule{0.167em}{0ex}}\text{mm}}$
Now, we can substitute the values into the work formula:
$W=\frac{1}{2}\left(\frac{{\epsilon }_{0}\pi {\left(\frac{6.0\phantom{\rule{0.167em}{0ex}}\text{cm}}{2}\right)}^{2}}{1.0\phantom{\rule{0.167em}{0ex}}\text{mm}}\right){\left(\frac{{Q}_{i}}{\frac{{\epsilon }_{0}\pi {\left(\frac{6.0\phantom{\rule{0.167em}{0ex}}\text{cm}}{2}\right)}^{2}}{2.0\phantom{\rule{0.167em}{0ex}}\text{mm}}}-\frac{{Q}_{i}}{\frac{{\epsilon }_{0}\pi {\left(\frac{6.0\phantom{\rule{0.167em}{0ex}}\text{cm}}{2}\right)}^{2}}{1.0\phantom{\rule{0.167em}{0ex}}\text{mm}}}\right)}^{2}$
Simplifying the expression, we have:
$W=\frac{1}{2}\left(\frac{{\epsilon }_{0}\pi {\left(\frac{6.0}{2}\right)}^{2}}{1.0}\right){\left(\frac{{Q}_{i}}{\frac{{\epsilon }_{0}\pi {\left(\frac{6.0}{2}\right)}^{2}}{2.0}}-\frac{{Q}_{i}}{\frac{{\epsilon }_{0}\pi {\left(\frac{6.0}{2}\right)}^{2}}{1.0}}\right)}^{2}\phantom{\rule{0.167em}{0ex}}\text{J}$

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