42 Gauss’s LawAdditional ProblemsP24.54In general,Eijk=++aybzcx±±±In the xyplane, z=0 andEik=+aycxΦΦEExwxwdaycxdAchxdxchxchw=⋅=+⋅===zzz==EAi kk±ej020222xyzx= 0x= wy = 0y = hdA= hdxFIG. P24.54P24.55(a)qQQQin−= +32(b)The charge distribution is spherically symmetric and qin>0 . Thus, the field is directedradially outward .(c)EkqrkQreein2for rc≥.(d)Since all points within this region are located inside conducting material,E=0forbrc<<.(e)ΦΦEEdq⇒=∈=z000in(f)in3(g)Erreein3(radially outward) for arb≤<.(h)qVQarQrain+FHGIKJFHGIKJρπ3433433333(i)ErkrQraraeee+FHGIKJ=in33333(radially outward) for 0≤≤ra.(j)From part (d), E=0 for . Thus, for aspherical gaussian surface with
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This note was uploaded on 12/14/2011 for the course PHY 203 taught by Professor Staff during the Fall '11 term at Indiana State University .