Up k 93 0 the page area is f k x 8z here 0 to the

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Unformatted text preview: ays F = 0 plate) B) da, so the force per 8z So 1m= PO0'2V2/2B. (up).! K = 93 0 the page. area is f = K x 8z Here 0 to the right, and of the lower is poO'v/2, into in general. to the right, and B (the field of the lower plate) is poO'v/2, into the page. So 1m= PO0'2V2/2 (up).! {X x2 {y {X y2 (c) The 5.23 (c) Wy electric field 2; Wz= Jo y' is Jo Zdx' 2" - force per unit area on the upper plate is Problem = Jo x'dx' = of the lower platedy'- O'/2Eo; the =electriczx. ak 1 k 1 1 Ie = O'2/2Eo (down). They balance(sk)z=-z; I/Eo,or v = y 1/..,ftOiIO =Z-- (the speed cp= c as in A",=k~B=VxA=-- as if POV2 s= J=-(VxB)=x as - )] of light), zcp. Prob. 5.12. s fLo fLo [ s fLoS 9.3 (10 points) Griffiths Problem 5.24 Problem 5.17 =yx+zy+xz=F. ,( 8/8x 8/8y 8/8z Problem 5.24 well (~-zx) the axes so the w~ point0 r lies on the (y2/2 - zx) (0, y, 0). Consider a source point .1 V x field x2/2 y axis: r = WeIw= ~Y+ orient might as at (x',y',z') on loop #1: V. = . (r x B) ~ = -~ [B. (V x r) -yr. (Vz;x dl' == 0,Xsinc...
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