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3 years ago

Select True or False for the follwing statements about electric field lines.

Physics

1 answer:

aalyn [17]3 years ago

5 0

Answer:

Please see below as the answer is self explanatory.

Explanation:

1) A positive point charge released from rest will initially accelerate along an E-field line : TRUE.

The electric field lines have the same direction that will have a positive test charge, when acted on by the force exerted by the charge distribution that creates the field.

2) E-Field lines make circles around positive charges: FALSE.

E-Field lines start in positive charges and finish in negative charges, except at infinity.

3) E-field lines point inward toward negative charges: TRUE

As the E-field lines follow the trajectory that would have a positive test charge, as negative charges attract the positive charges, the positive test charge will travel towards the negative charges.

4) E-Field line may cross: FALSE

As the E-Field is the gradient of the electric potential (so they have the direction of the maximum change of the potential), if they may cross, this would mean that the potential could be changing differently at the same point, which it is not possible.

5) E-Field lines point outward from positive charges: TRUE

As explained above, the E-field lines follow the trajectory that it would have a positive test charge, so if the field is created by a positive charge, the test charge would go away from it, so the field lines point outward from positive charges.

6) Where the E-field lines are dense the E-field must be weak : FALSE

By convention, the E-field lines are proportional to the charge that creates the field, so if the E-field lines are dense, this means that the field is strong.

7) E-field lines do not begin or end in a charge-free region except at infinity: TRUE.

Except at infinity, E-field lines originate at positive charges and finish at negative charges, so they can't originate in a charge-free region.

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Answer:

Given:

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$k_{b} = 1.38\times 10^{- 23} J/k$ = 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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Answer:

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