problem22_36

University Physics with Modern Physics with Mastering Physics (11th Edition)

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  • PresidentHackerCaribou10582
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22.36: a) For , 0 , = < E a r since no charge is enclosed. For , , 2 0 4 1 r q πε E b r a = < < since there is + q inside a radius r . For = < < E c r b , 0, since now the – q cancels the inner + q . For , , 2 0 4 1 r q πε E c r = since again the total charge enclosed is +
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  • '
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  • Interior, total charge, inner shell surface, outer shell surface

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