46 Gauss’s LawP24.63The magnitude of the field due to the each sheet given byEquation 24.8 isE=∈σ20directed perpendicular to the sheet.(a)In the region to the left of the pair of sheets, both fields aredirected toward the left and the net field isFIG. P24.63E=∈0to the left .(b)In the region between the sheets, the fields due to the individual sheets are oppositelydirected and the net field isE=0.(c)In the region to the right of the pair of sheets, both are fields are directed toward the rightand the net field isE=∈0to the right .P24.64The resultant field within the cavity is the superposition of twofields, one E+due to a uniform sphere of positive charge of radius2a, and the other E−due to a sphere of negative charge of radius acentered within the cavity.434302πρπrrE∈FHGIKJ=+soErr+=∈=∈ρr3300±–−∈F
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Electric charge, charge density, net field, 46