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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 Electric charge, 15 m, Gaussian sphere

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