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timofeeve [1]
1 year ago
6

A uniformly charged, straight filament 7.00m in length has a total positive charge of 2.00μC. An uncharged card-board cylinder 2

.00cm in length and 10.0cm in radius surrounds the filament at its center, with the filament as the axis of the cylinder. Using reasonable approximations, find (a) the electric field at the surface of the cylinder
Physics
1 answer:
TEA [102]1 year ago
4 0

The electric field at the surface of the cylinder is 51428V/m

Given data:

• The length of the charge is l=  7m.

• The charge is q = 2μC..

• The radius the cylinder is r = 10 cm

Since the filament length is so large as compared to the cylinder length that the infinite line of charge can be assumed.

The expression to calculate the electric field is given as,

E=2kλ/r

Here, λ is the linear charge density.

Substitute the values in the above equation,

E = (2×9×109N⋅m^2/C^2×2×10^−6C)/0.1m×7m

E = 51428N/C×(V/m)/(N/C)

=51428V/m

An electric charge is the property of matter where it has more or fewer electrons than protons in its atoms. Electrons carry a negative charge and protons carry a positive charge. Matter is positively charged if it contains more protons than electrons, and negatively charged if it contains more electrons than protons.

Learn more about charge here:

brainly.com/question/19886264

#SPJ4

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An electron and a proton each have a thermal kinetic energy of 3kBT/2. Calculate the de Broglie wavelength of each particle at a
S_A_V [24]

Answer:

Given:

Thermal Kinetic Energy of an electron, KE_{t} = \frac{3}{2}k_{b}T

k_{b} = 1.38\times 10^{- 23} J/k = Boltzmann's constant

Temperature, T = 1800 K

Solution:

Now, to calculate the de-Broglie wavelength of the electron, \lambda_{e}:

\lambda_{e} = \frac{h}{p_{e}}

\lambda_{e} = \frac{h}{m_{e}{v_{e}}              (1)

where

h = Planck's constant = 6.626\times 10^{- 34}m^{2}kg/s

p_{e} = momentum of an electron

v_{e} = velocity of an electron

m_{e} = 9.1\times 10_{- 31} kg = mass of electon

Now,

Kinetic energy of an electron = thermal kinetic energy

\frac{1}{2}m_{e}v_{e}^{2} = \frac{3}{2}k_{b}T

}v_{e} = \sqrt{2\frac{\frac{3}{2}k_{b}T}{m_{e}}}

}v_{e} = \sqrt{\frac{3\times 1.38\times 10^{- 23}\times 1800}{9.1\times 10_{- 31}}}

v_{e} = 2.86\times 10^{5} m/s                    (2)

Using eqn (2) in (1):

\lambda_{e} = \frac{6.626\times 10^{- 34}}{9.1\times 10_{- 31}\times 2.86\times 10^{5}} = 2.55 nm

Now, to calculate the de-Broglie wavelength of proton, \lambda_{e}:

\lambda_{p} = \frac{h}{p_{p}}

\lambda_{p} = \frac{h}{m_{p}{v_{p}}                             (3)

where

m_{p} = 1.6726\times 10_{- 27} kg = mass of proton

v_{p} = velocity of an proton

Now,

Kinetic energy of a proton = thermal kinetic energy

\frac{1}{2}m_{p}v_{p}^{2} = \frac{3}{2}k_{b}T

}v_{p} = \sqrt{2\frac{\frac{3}{2}k_{b}T}{m_{p}}}

}v_{p} = \sqrt{\frac{3\times 1.38\times 10^{- 23}\times 1800}{1.6726\times 10_{- 27}}}

v_{p} = 6.674\times 10^{3} m/s                               (4)                    

Using eqn (4) in (3):

\lambda_{p} = \frac{6.626\times 10^{- 34}}{1.6726\times 10_{- 27}\times 6.674\times 10^{3}} = 5.94\times 10^{- 11} m = 0.0594 nm

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Types of media that can influence one's social development include all of the following except
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An 80.0 kg hiker walks a distance of 400.0 m along a road that slopes 5.0 degrees upward, and then stops. What is the hiker's fi
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vodomira [7]

Answer:

The force of the impact would be smaller

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If the force is big then the time would be small (2500N of Force = 10 seconds)

If the force is small then the time would be big (250N of Force = 50 seconds)

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