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?

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

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:

$C=\frac{{\u03f5}_{0}A}{d}$

A capacitor consists of circular plates, so its surface is:

$A=\pi \times {r}^{2}$

And radius is:

$r=\frac{D}{2}=\frac{6}{2}=3cm$

$\u03f5}_{0$ is the dialectric constznt of vacuum and has the following value:

$\u03f5}_{0}=8.854\times {10}^{-12}\frac{F}{m$

Thus, the first formula becomes:

$U=\frac{1}{2}\times \frac{{\u03f5}_{0}\pi {r}^{2}}{d}\times {V}^{2}$

Now, convert cm and mm into m

Finally, we have:

$U=\frac{1}{2}\times \frac{8.854\times {10}^{-12}\pi \times {(3\times {10}^{-2})}^{2}}{1\times {10}^{-3}}\times {\left(150\right)}^{2}$

$U=0.28\times {10}^{-6}J$

b)Here the distance between the plates increases to${d}_{1}=2mm$

The work required to achieve will be obtained as the energy difference:

$W=U-{U}_{1}$

${U}_{1}=\frac{1}{2}\times \frac{8.854\times {10}^{-12}\pi \times {(3\times {10}^{-2})}^{2}}{2\times {10}^{-3}}\times {150}^{2}=0.14\times {10}^{-6}J$

And we have:

$W=0.28\times {10}^{-6}-0.14\times {10}^{-6}$

$W=0.14\times {10}^{-6}J$

a) The energy stored in the capacitor will be obtained using this formula:

Where

The capacitance we can get from the following formula:

A capacitor consists of circular plates, so its surface is:

And radius is:

Thus, the first formula becomes:

Now, convert cm and mm into m

Finally, we have:

b)Here the distance between the plates increases to

The work required to achieve will be obtained as the energy difference:

And we have:

Nick Camelot

Skilled2023-06-19Added 164 answers

Eliza Beth13

Skilled2023-06-19Added 130 answers

Mr Solver

Skilled2023-06-19Added 147 answers

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