Chapter 2439P24.40From Gauss’s Law,EAQ=∈0σ==∈=×− =− ×=−−−QAE012988510130110115...ejafCmnCm22P24.41The fields are equal. The Equation 24.9 E=∈conductor0for the field outside the aluminum looksdifferent from Equation 24.8 E=∈insulator20for the field around glass. But its charge will spread out tocover both sides of the aluminum plate, so the density is conductor=QA2. The glass carries chargeonly on area A, with insulator=QA. The two fields are QA20∈the same in magnitude, and both areperpendicular to the plates, vertically upward if Qis positive.*P24.42(a)All of the charge sits on the surface of the copper sphere at radius 15 cm. The field inside iszero .(b)The charged sphere creates field at exterior points as if it were a point charge at the center:E××=×−kqre29924899104010017124 10awayNmCC moutwardN C outward22...ejaf(c)
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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 .