36 Gauss’s LawP24.28σ=×FHGIKJ−−860 10100622..CcmcmmCm22ejE=∈=××−−22 8 85 10486 1002129...N C away from the wallThe field is essentially uniform as long as the distance from the center of the wall to the field point ismuch less than the dimensions of the wall.P24.29If ρis positive, the field must be radially outward. Choose as thegaussian surface a cylinder of length Land radius r, contained insidethe charged rod. Its volume is πrL2and it encloses charge ρπ2.Because the charge distribution is long, no electric flux passesthrough the circular end caps; EA⋅=°=dEdAcos.90 00 . The curvedsurface has °dAcos0 , and Emust be the same strengtheverywhere over the curved surface.FIG. P24.29Gauss’s law, ∈zdq0,becomesEdACurvedSurfacez=∈20.Now the lateral surface area of the cylinder is 2rL:ErL220bg=∈.Thus,E=∈r20radially away from the cylinder axis .*P24.30Let represent the charge density. For the field inside the sphere at r15=cm we haveqr1120130443πρ=∈=∈insideEr1103=∈=∈=×−×=−×−−33 8 85 1086 10005457 10
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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 .