699_Physics ProblemsTechnical Physics

# 699_Physics ProblemsTechnical Physics - Chapter 24 P24.40 =...

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Chapter 24 39 P24.40 From Gauss’s Law, EA Q = 0 σ == ∈= × − = − × = −− Q A E 0 12 9 8 8 51 0 1 3 0 1 1 0 1 1 5 ... ej a f Cm nCm 22 P24.41 The fields are equal. The Equation 24.9 E = conductor 0 for the field outside the aluminum looks different from Equation 24.8 E = insulator 2 0 for the field around glass. But its charge will spread out to cover both sides of the aluminum plate, so the density is conductor = Q A 2 . The glass carries charge only on area A , with insulator = Q A . The two fields are Q A 2 0 the same in magnitude, and both are perpendicular to the plates, vertically upward if Q is positive. *P24.42 (a) All of the charge sits on the surface of the copper sphere at radius 15 cm. The field inside is zero . (b) The charged sphere creates field at exterior points as if it were a point charge at the center: E ×× kq r e 2 99 2 4 8 9 91 0 4 01 0 017 124 10 away Nm C C m outward N C outward 2 2 . . . e j af (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 .

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