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3 years ago

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An ideal parallel-plate capacitor having circular plates of diameter D that are a distance d apart stores energy U when it is co

nnected across a fixed potential difference. If you want to triple the amount of energy stored in this capacitor by changing only the size of its plates, the diameter should be changed to______________.

1 answer:

Stella [2.4K]3 years ago

3 0

Answer: $\sqrt{3}D$

Explanation:

Given

Diameter of Plates is D

distance between Plates is d

Energy Stored in the Plates is U

Suppose fixed Potential is V

Energy is given by

$U=\frac{1}{2}CV^2$

where C=capacitance

$C=\frac{\epsilon A}{d}$

where A=area of Plates

$A=\frac{\pi D^2}{4}$

Thus $C=\frac{\epsilon \cdot \pi \cdot D^2}{4d}$

Keeping all other factors as same

$U\propto D^2----1$

For triple the Energy

$3U\propto (D')^2-----2$

divide 1 and 2 we get

$\frac{U}{3U}=\frac{D^2}{D'^2}$

$D'=\sqrt{3}D$

Thus the diameter should be change to $\sqrt{3}$ times of original value

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Before the engines fail, the rocket's altitude at time <em>t</em> is given by

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Solve for <em>t</em> to find this time to be

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At this time, the rocket attains a velocity of

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When it's in freefall, the rocket's altitude is given by

$y_2(t)=1150\,\mathrm m+\left(124\dfrac{\rm m}{\rm s}\right)t-\dfrac g2t^2$

where $g=9.80\frac{\rm m}{\mathrm s^2}$ is the acceleration due to gravity, and its velocity is

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So the rocket is in motion for a total of 11.2 s + 32.6 s = 43.4 s.

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Solve for $y_{\rm max}$ and we find that the rocket reaches a maximum altitude of about 1930 m.

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