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Unformatted text preview: 43. At all points where there is an electric ﬁeld, it is radially outward. For each part of the problem, use a Gaussian surface in the form of a sphere that is concentric with the sphere of charge and passes through the point where the electric ﬁeld is to be found. The ﬁeld is uniform on the surface, so E · dA = 4πr 2 E where r is the radius of the Gaussian surface. (a) Here r is less than a and the charge enclosed by the Gaussian surface is q (r/a)3 . Gauss’ law yields 4πr2 E = q ε0 r a
3 =⇒ E = qr . 4πε0 a3 (b) In this case, r is greater than a but less than b. The charge enclosed by the Gaussian surface is q , so Gauss’ law leads to q q 4πr2 E = =⇒ E = . ε0 4πε0 r2 (c) The shell is conducting, so the electric ﬁeld inside it is zero. (d) For r > c, the charge enclosed by the Gaussian surface is zero (charge q is inside the shell cavity and charge −q is on the shell). Gauss’ law yields 4πr2 E = 0 =⇒ E = 0 . (e) Consider a Gaussian surface that lies completely within the conducting shell. Since the electric ﬁeld is everywhere zero on the surface, E · dA = 0 and, according to Gauss’ law, the net charge enclosed by the surface is zero. If Qi is the charge on the inner surface of the shell, then q + Qi = 0 and Qi = −q . Let Qo be the charge on the outer surface of the shell. Since the net charge on the shell is −q , Qi + Qo = −q . This means Qo = −q − Qi = −q − (−q ) = 0. ...
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This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.
 Fall '08
 SPRUNGER
 Physics, Charge

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