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irga5000 [103]
3 years ago
6

It is known that the gravitational force of attraction between two alpha particles is much weaker than the electrical repulsion.

For two alpha particles at a distance d apart, calculate the ratio of the size of the gravitational attraction to that of the electrical repulsion. Specifically, find the magnitude of Gravitational/Electrical.
Physics
1 answer:
natali 33 [55]3 years ago
8 0

Answer:

<em>The ratio of gravitational force to electrical force is 3.19 x 10^-36 </em>

<em></em>

Explanation:

mass of an alpha particle = 6.64 x 10^{-27} kg

charge on an alpha particle = +2e = +2(1.6 x 10^{-19} C) = 3.2 x 10^{-19} C

distance between particles = d

For gravitational attraction:

The force of gravitational attraction F = \frac{Gm^{2} }{r^{2} }

where G = gravitational constant = 6.67 x 10^{-11} m^3 kg^-1 s^-2

r = the distance between the particles = d

m = the mass of each particle

therefore, gravitational force = \frac{6.67*10^{-11}*(6.64*10^{-27} )^{2}  }{d^{2} } = \frac{2.94*10^{-63} }{d^{2} }  Newton

For electrical repulsion:

Electrical force between the particles = \frac{-kQ^{2} }{r^{2} }

where k is the Coulomb's constant = 9.0 x 10^{9} N•m^2/C^2

r = distance between the particles = d

Q = charge on each particle

therefore, electrical force = \frac{-9*10^{9}*(3.2*10^{-19} )^{2}  }{d^{2} } = \frac{-9.216*10^{-28} }{d^{2} } Newton

the negative sign implies that there is a repulsion on the particles due to their like charges.

Ratio of the magnitude of gravitation to electrical force = \frac{2.94*10^{-63} }{9.216*10^{-28} }

==> <em>3.19 x 10^-36 </em>

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Answer:

a. B= 9.45 \times10^{-3} T

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First, look at the picture to understand the problem before to solve it.

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B=\frac{u_{0}I_{w}}{2\pi d_{1}} =\frac{52 \times4\pi \times10^{-7} }{2\pi 1.1 \times 10^{-3}} =9.45 \times10^{-3} T\\

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b. d2=3.6 mm

Here, the point is located in the surface of the cilinder. Therefore, we have to consider the current density of the conductor to calculate the magnetic field as follows:

J: current density

c: outer radius

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J= \frac {-I_{c}}{\pi(c^{2}-b^{2}  ) }}

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c. d3=7.4 mm

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B=\frac{u_{0}(I_w-I_c)}{2\pi d_3 } =\frac{2.011\times10^-5}{3.441\times10^{-4}} =0.0584 T

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