696_Physics ProblemsTechnical Physics

# 696_Physics ProblemsTechnical Physics - 36 Gausss Law...

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36 Gauss’s Law P24.28 σ F H G I K J −− 860 10 100 6 2 2 .. Ccm cm m Cm 22 ej E = = × × 2 2 8 85 10 486 10 0 2 12 9 . . . N C away from the wall The field is essentially uniform as long as the distance from the center of the wall to the field point is much less than the dimensions of the wall. P24.29 If ρ is positive, the field must be radially outward. Choose as the gaussian surface a cylinder of length L and radius r , contained inside the charged rod. Its volume is π rL 2 and it encloses charge ρπ 2 . Because the charge distribution is long, no electric flux passes through the circular end caps; EA ⋅= ° = dE d A cos . 90 0 0 . The curved surface has ° d A cos0 , and E must be the same strength everywhere over the curved surface. FIG. P24.29 Gauss’s law, z d q 0 , becomes Ed A Curved Surface z = 2 0 . Now the lateral surface area of the cylinder is 2 rL : Er L 2 2 0 bg = . Thus, E = r 2 0 radially away from the cylinder axis . *P24.30 Let represent the charge density. For the field inside the sphere at r 1 5 = cm we have qr 11 2 0 1 3 0 4 4 3 πρ = = inside E r 1 1 0 3 = = = ×− × =− × 3 3 8 85 10 86 10 005 457 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 .

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