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- Thread starter crazy student
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Andrew Mason

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What is the potential energy of a charge q1 a distance r from another charge q2? What is the potential energy of a proton whose surface is situated 3.6E-15 m. from the centre of a silicon nucleus?crazy student said:

AM

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I think it suppose r starts from infinity

but i can't get to the model ans 4.2E-6 J

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OlderDan

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You need to check the exponent in that answer, and you need to interpret "just penetrate" as the outer surface of the proton just barely penetrating the outer surface of the silicon nucleus.crazy student said:

I think it suppose r starts from infinity

but i can't get to the model ans 4.2E-6 J

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I found the question quite confusing also....

from the model ans and equation V = KQ/r

r = 4.8E-15 m, which is the sum of proton and silicon nucleus radii

but i don't know the reason behind....

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Andrew Mason

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The force between charges depends on the distance from the centre of the charge. The electric field of a charge is given by Gauss' law, which states that the field depends only on the enclosed charge:crazy student said:

I found the question quite confusing also....

from the model ans and equation V = KQ/r

r = 4.8E-15 m, which is the sum of proton and silicon nucleus radii

but i don't know the reason behind....

[tex]\int_S E\cdot dA = \frac{q_{encl}}{\epsilon_0}[/tex]

For a sphere:

[tex]\int_S E\cdot dA = 4\pi r^2E = \frac{q_{encl}}{\epsilon_0}[/tex]

[tex]E = \frac{q_{encl}}{4\pi \epsilon_0 r^2} = \frac{kq_{encl}}{r^2}[/tex]

So if you assume that a proton and a silicon nucleus are uniform spheres, the field at their surface depends only upon the distance of that surface from the respective centres.

AM

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