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Elis [28]
3 years ago
5

Consider a long cylindrical charge distribution of radius R = 17 cm with a uniform charge density of rho = 15 C/m3. Find the ele

ctric field at a distance r = 26 cm from the axis.
Physics
1 answer:
zzz [600]3 years ago
5 0

Answer:

E = 9.4*10^{10}N/C

Explanation:

We use Gauss's law which says

$\int E\cdot dA = \frac{Q_{enc}}{\varepsilon}. $     (1)

Now, for the cylindrical charge distribution the charge enclosed is

Q_{enc} = \rho V

where V is the volume of the cylinder.

To evaluate Gauss's law, the Gaussian surface we choose is a cylinder concentric with the charged cylinder; therefore, equation  (1) becomes

E (2\pi rL )=\dfrac{\rho V}{\varepsilon _o}

E (2\pi rL )=\dfrac{\rho \pi R^2L}{\varepsilon _o}

E =\dfrac{\rho \pi R^2L}{ (2\pi rL )\varepsilon _o }

\boxed{E =\dfrac{\rho R^2}{ 2\varepsilon _o r }}

Putting in numerical values

\rho = 15C/m^3

R = 17cm =0.17m

r = 26cm=0.26m

\varepsilon_0 =8.85*10^{-12}m^{-3}kg^{-1}s^4A^2}

we get:

E =\dfrac{15 (0.17)^2}{ 2(8.85*10^{-12}) (0.26) }

\boxed{E = 9.4*10^{10}N/C}

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4 0
3 years ago
A positive kaon (K+) has a rest mass of 494 MeV/c² , whereas a proton has a rest mass of 938 MeV/c². If a kaon has a total energ
vitfil [10]

Answer:

<em>0.85c </em>

Explanation:

Rest mass of Kaon M_{0K} = 494 MeV/c²

Rest mass of proton M_{0P}  = 938 MeV/c²

The rest energy is gotten by multiplying the rest mass by the square of the speed of light c²

for the kaon, rest energy E_{0K} = 494c² MeV

for the proton, rest energy E_{0P} = 938c² MeV

Recall that the rest energy, and the total energy are related by..

E = γE_{0}

which can be written in this case as

E_{K} = γE_{0K} ...... equ 1

where E = total energy of the kaon, and

E_{0} = rest energy of the kaon

γ = relativistic factor = \frac{1}{\sqrt{1 - \beta ^{2} } }

where \beta = \frac{v}{c}

But, it is stated that the total energy of the kaon is equal to the rest mass of the proton or its equivalent rest energy, therefore...

E_{K} = E_{0P} ......equ 2

where E_{K} is the total energy of the kaon, and

E_{0P} is the rest energy of the proton.

From E_{K} = E_{0P} = 938c²    

equ 1 becomes

938c² = γ494c²

γ = 938c²/494c² = 1.89

γ = \frac{1}{\sqrt{1 - \beta ^{2} } } = 1.89

1.89\sqrt{1 - \beta ^{2} } = 1

squaring both sides, we get

3.57( 1 - \beta^{2}) = 1

3.57 - 3.57\beta^{2} = 1

2.57 = 3.57\beta^{2}

\beta^{2} = 2.57/3.57 = 0.72

\beta = \sqrt{0.72} = 0.85

but, \beta = \frac{v}{c}

v/c = 0.85

v = <em>0.85c </em>

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3 years ago
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Answer:

i think the anwer is C

Explanation:

6 0
3 years ago
What do you need to make 100 minutes charges neutral
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HELP PLEASE I need to finish this asap
ELEN [110]

Answer:

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Explanation:

pls vote brainliest

6 0
2 years ago
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