quiz_solutions02

Fundamentals of Physics, (Chapters 21- 44) (Volume 2)

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PY212 Solutions Quiz 2 We approximate the nucleus with a sphere of radius R = 6.64 fm = 6.64×10 –15 m filled with a uniform charge of Q = 94 e = 1.5×10 –17 C. This charge is positive, so the electric field will be directed radially outwards . Since we have a spherical symmetry in this problem, we can use Gauss’ law to find the electric field E : 0 enc tot ε q a d E = = Φ r v , Here the integral is over the surface area of a spherical Gaussian surface with arbitrary radius r . We should place the center of our Gaussian sphere at the center of our charged sphere, such that (1) the field E v is everywhere perpendicular to the Gaussian surface, = da E a d E r v , and (2) the magnitude of E is the same everywhere on the Gaussian sphere, ) 4 ( 2 r E EA da E da E π = = = , where A is the area of the Gaussian sphere (not that of the charged sphere!). Hence, the magnitude of the electric field at every point in space is given by
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Unformatted text preview: . 4 2 enc r q E πε = Note that we have two separate regions; inside the charged sphere ( r < R ) and outside the charged sphere ( r > R ). For the latter case the Gaussian surface encloses all the charge, i.e. 2 2 7 2 1 ) /C m N 10 35 . 1 ( 4 r r Q E ⋅ × = = − (for r > R ; outside). To determine the field inside the charged sphere, we need to determine the amount of charge enclosed by the Gaussian sphere when r < R . It is equal to V q ρ = enc where V = (4/3) πr 3 is the volume of the Gaussian sphere, and ρ is the charge density: 3 3 3 3 enc 3 sphere charged ) 3 / 4 ( ) 3 / 4 ( ) 3 / 4 ( r R Q r R Q q R Q V Q = = ⇒ = = Hence, r R r Qr r q E ) C) N/(m 10 61 . 4 ( 4 4 35 3 2 3 2 enc ⋅ × = = = (for r < R ; inside). Note that at r = R both relations give E = 3.06×10 21 N/C....
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This document was uploaded on 07/10/2009.

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