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# j0101 - Jackson Problem 1.1 B J Mattson 1 Part a I’m...

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Unformatted text preview: Jackson Problem 1.1 B. J. Mattson 1 Part a I’m assuming that we are dealing with a conductor at equilibrium. Then when we place excess charge on it. Consider a Gaussian surface with an enclosed charge of q’. In this case: —. 1 /E ' ﬁx da : _Qenclosed 60 or 6 7E 0. Since charges are free to move inside a conductor, the excess charge will move to follow the ﬁeld. But then we are not talking about a conductor in equilibrium. :><= So, the charge enclosed in the Gaussian volume must be zero. But the volume chosen was arbitrary, so stretch it just to the edge of the conductor and the enclosed charge must still be zero. the only place left for the excess charge is the surface. 2 Part b First consider such a conductor with interior charges. Total charge on the original con— ductor is Q; charge placed in the interior is q. Take a Guassian surface just outside the surface. Then, fE-ﬁzl (:EdV (2) 60 V So, ELA2Q+Q (3) 60 Q+q E = 4 J— AGO And E L 3E 0, so the outside is not protected from those interior charges. Now consider a conductor with external charges. Take a path from point a on the surface to point b on the surface straight through the conductor. Then use: jiE-Jzzo (5) So, / E - d7 E - dl = 0 (6) surfacepath interiorpath But the integral over the surface part of the path is 0, since there is no ﬁeld arbitrarily close to the surface. Then —a —; f E ~ cl! = 0 (7) interiorpath But this means that, if Emterior 7Q 0, Va ¢ Vb (8) However, a conductor is an equipotential. Therefore, the elecric ﬁeld on the interior of the sphere must be zero. 3 Part 0 Consider a conductor, and a path enclosing a bit of the surface. Now, we know that: fE-dizo (9) Now, squish the side of the path which are perpendicular to the surface of the conduc- tor; then their contribution to the above integral is zero. So, fE-cizzf Edi—f E-CZZ (10) inside outside But E = 0 inside the conductor, so yiE-cflzf E-cilzo (11) outside Or, Eloutsidel = 0 Since l is arbitrary, E Loutside = 0. Therefore, the electric ﬁeld just outside of a con— ductor is perpendicular to the surface of the conducotr. Now take a Gaussiand pill box enclosing the surface. Squish the side walls, so that they are inﬁmitesimal. Then, —. 1 f E - a = — pm d3a: (13) S 60 V can be written: —I A —b A 1 _‘ 3 f E-n+ E-n=— p(\$)d\$ (14) inside outside 60 V But Emside = 0. NOW squish the box so that E N constant over A. Then, —. A 74 E - 71 = _‘7 (15) outside 60 Or, A Eoutside A = —0 60 _ (7 And Eoutside — _ ...
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j0101 - Jackson Problem 1.1 B J Mattson 1 Part a I’m...

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