HW9_sol

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

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online