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CourseNotes.12 - Figure 8 Charged shell with a positive...

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Figure 8: Charged shell with a positive point charge at its center. by our Gaussian surface of radius r includes the central point charge as well as the charge from the enclosed portion of the shell. Lets first calculate the charge from the shell enclosed by the Gaussian sphere. This is not a simple as multiplying the charge density times the volume because the charge density is a funtion of r . We must therefore sum up all of the contributions at each value of r (i.e. integrate). Q enc,shell = Z ρ dV = Z r a A r 0 4 πr 0 2 dr 0 = 4 πA Z r a r 0 dr 0 = 2 πA ( r 2 - a 2 ) Plugging this into Gauss’s law and exploiting the symmetry gives the electric field with in the shell. E = Q enc 0 A = 1 0 4 πr 2 q + 2 πA ( r 2 - a 2 ) = A 2 0 + q - 2 πAa 2 4 π 0 r 2 The question at hand is what does the value of A need to be such that E is independent of r ? Clearly, the first term is a constant, but the second term has some r dependence. In order for E to be a constant, the numerator in the second term must therefore vanish.
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