674_Physics ProblemsTechnical Physics

674_Physics ProblemsTechnical Physics - 14 Electric Fields...

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14 Electric Fields P23.35 (a) The electric field at point P due to each element of length dx , is dE kdq xy e = + 22 and is directed along the line joining the element to point P . By symmetry, Ed E xx == z 0 and since dq dx = λ , E E dE dE yy = zz cos θ where cos = + y . Therefore, Ek y dx k y e e = + = z 2 2 32 0 2 0 λθ ej A sin . FIG. P23.35 (b) For a bar of infinite length, 0 90 and E k y y e = 2 . P23.36 (a) The whole surface area of the cylinder is Ar r Lr r L =+= + 222 2 πππ af . QA × + = × σπ 15 0 10 2 0 025 0 0 025 0 0 060 0 2 00 10 9 10 .. . . . Cm m m m C 2 bg (b) For the curved lateral surface only, L = 2 π . × = × −− 15 0 10 2 0 025 0 0 060 1 41 10 91 0 . . m 0 m C 2 (c) QV r L = × = × ρρ 29 2 11 500 10 0 025 0 0 060 0 5 89 10 m m C 3 . P23.37 (a) Every object has the same volume, V × 8 0 030 2 16 10 3 4 0 m m 3 . For each, × × = × ρ 400 10 2 16 10 8 64 10 94 1 1 m C 33 e j (b) We must count the 900 . c m 2 squares painted with charge: (i) 642 4 ×= squares × × = × σ 15 0 10 24 0 9 00 10 3 24 10 1 0 . . m C e j (ii) 34 squares exposed × × = × 15 0 10 34 0 9 00 10
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