706_Physics ProblemsTechnical Physics

706_Physics ProblemsTechnical Physics - 46 Gausss Law...

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46 Gauss’s Law P24.63 The magnitude of the field due to the each sheet given by Equation 24.8 is E = σ 2 0 directed perpendicular to the sheet. (a) In the region to the left of the pair of sheets, both fields are directed toward the left and the net field is FIG. P24.63 E = 0 to the left . (b) In the region between the sheets, the fields due to the individual sheets are oppositely directed and the net field is E = 0 . (c) In the region to the right of the pair of sheets, both are fields are directed toward the right and the net field is E = 0 to the right . P24.64 The resultant field within the cavity is the superposition of two fields, one E + due to a uniform sphere of positive charge of radius 2 a , and the other E due to a sphere of negative charge of radius a centered within the cavity. 4 3 4 3 0 2 πρ π r rE F H G I K J = + so Er r + = = ρ r 33 00 ± F
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