Flux and nonconducting shells. A charged particle is suspended at the center of
ID: 1686826 • Letter: F
Question
Flux and nonconducting shells. A charged particle is suspended at the center of two concentric spherical shells that are very thin and made of nonconducting material. Figure (a) shows a cross section. Figure (b) gives the net flux through a Gaussian sphere centered on the particle, as a function of the radius r of the sphere. (The vertical axis is marked in increments of 1.0 106 N·m2/C with Fs = 5.0 106N·m2/C )Explanation / Answer
From Gauss's Law for net charge "Q" enclosed byclosed surface "S": {Net E Flux ThruClosed Bounding Surface "S"} = F = = {Net Enclosed Charge Q}/e0 ----> Q = e0*F = (8.854e-12)*F Item(a): Using Figure 23-31 and applying Gauss's Law to a spherical surface inside shell "A" which encloses central particle charge"QP": QP = (8.854e-12)*F = (8.854e-12)*(2.0e+5) = 1.771e-6C = 1.771 µC Item (b): Using Figure 23-31 and applying Gauss's Law to a spherical surface between shells "A" & "B" which encloses central particle charge"QP" and shell "A" net charge "QA": {QA + QP} = (8.854e-12)*F ----> QA = (8.854e-12)*F - QP = (8.854e-12)*(-4.0e+5) - (1.771e-6) = -5.3126e-6 C = -5.3126µC Item (c): Using Figure 23-31 and applying Gauss's Law to a spherical surface outside shells "A" & "B" which encloses central particle charge"QP", shell "A" net charge "QA", and shell "B" net charge"QB": {QB +QA + QP} = (8.854e-12)*F ----> QB = (8.854e-12)*F - QP -QA = (8.854e-12)*(6.0e+5) -(1.771e-6)- (-5.3126e-6) = 8.854e-6 C = 8.854 µC