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ANTONII [103]
3 years ago
10

A capacitor is not the most efficient device for storing energy. Batteries can store more energy in much less space. For example

, a typical 12 V automobile battery stores on the order of 1.00 x 10^6 J. (a) Find the capacitance necessary to store 1.00 x 10^6 J with a potential difference of 1.00 x 10^4 V across the capacitor's terminals.
(b) Suppose that such a capacitor was made in the form of a parallel-plate capacitor with a vacuum between the plates and an electric field no greater than 9.00 x 10^6 V/m. What is the minimum area of the plates?
Physics
1 answer:
photoshop1234 [79]3 years ago
5 0

Answer:

A=2.49\times 10^6\ m^2

Explanation:

Given that

Stored energy E

U=10^6\ J

a)

We know that stored energy in capacitor given as

U=\dfrac{1}{2}CV^2

Given that

V=10^4\ V

U=\dfrac{1}{2}CV^2

10^6=\dfrac{1}{2}\times C\times (10^4)^2

C= 0.02 F

b)

Electric filed E = 9 x 10^6 V/m

We know that

V = E .d

10^4=9\times 10^6\times d

d=1.11 mm

We know that

C=\dfrac{\varepsilon _oA}{d}

0.02=\dfrac{8.89\times 10^{-12}A}{1.11\times 10^{-3}}

A=2.49\times 10^6\ m^2

This is area  area of the plates.

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Explanation:

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A lead ball is dropped into a lake from a diving board 5.0 m above the water. After entering the water, it sinks to the bottom w
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Answer:

|D_{depth} |=19.697m

Explanation:

To find Depth D of lake we must need to find the time taken to hit the water.So we use equation of simple motion as:

Δx=vit+(1/2)at²

x_{f}-x_{i}=v_{i}t+(1/2)at^{2}\\  -5.0m=(o)t+(1/2)(-9.8m/s^{2} )t^{2}\\ -4.9t^{2}=-5.0\\ t^{2}=5/4.9\\t=\sqrt{1.02} \\t=1.01s

As we have find the time taken now we need to find the final velocity vf from below equation as

v_{f}=v_{i}+at\\v_{f}=0+(-9.8m/s^{2} )(1.01s) \\v_{f}=-9.898m/s

So the depth of lake is given by:

first we need to find total time as

t=3.0-1.01 =1.99 s

|D_{depth} |=|vt|\\|D_{depth} |=|(-9.898m/s)(1.99s)|\\|D_{depth} |=19.697m

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