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CH24 - Q24.7 Faraday’s visualization of electric field...

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Unformatted text preview: Q24.7 Faraday’s visualization of electric field lines lends insight to this question. Consider a section of a vertical sheet carrying charge +1 coulomb. It has 3- field lines pointing out frorn it horizontally to ~- 60 the right and left, all uniformly spaced. The lines have the same uniform spacing close to the sheet and far away, showing that the field has the same value at all distances. (224.9 (224.10 Inject some charge at arbitrary places within a conducting object. Every bit of the charge repels every other bit, so each bit runs away as far as it can, stopping only when it reaches the outer surface of the conductor. If the person is uncharged, the electric field inside the sphere is zero. The interior wall of the shell carries no charge. The person is not harmed by touching this wall. If the person carries a (small) charge q, the electric field inside the sphere is no longer zero. Charge —q is induced on the inner wall of the sphere. The person will get a (small) shock when touching the sphere, as all the charge on his body jumps to the metaL (12.4.13 Gauss’s law predicts, as described in section 24.4, that excess charge on a conductor will reside on the surface of the conductor. If a car is left charged by a lightning strike, then that charge will remain on the outside of the car, not banning the occupants. It turns out that during the lightning strike, the current also remains on the outside of the conductor. Note that it is not necessarily safe to be in a fiberglass car or a convertible during a thunderstorm. 199 924.24 (a) E “3 E II II k‘Q,_ (8. 99): 109)(26. o x 18*6 )(8.100)_ (b) E a3 (0. 400)3 ll 1:20 (8..99x109](260x10‘5) (‘3 E=T=W=- 3' k¢_Q_ [8..99x109)(260x10'6) (d) E: —2 =—(0:W—=— f The direction for each electric field is W. 100cm 2 P2428 a:(8.60x10_6 C/cm3)[ ] =8.60><10’2 C/m3 _ a _ 8.60 x 10-2 2 so 2(335 x 10'”) The field is essentially uniform as long as the distance from the center of the wall to the field pointi much less than the dimensions of the wall. m E = 4.86 x 109 N/C away from the wall P2429 If p is positive, the field must be radially outward. Choose as the gaussian surface a cylinder of length L and radius 7, contained inside the charged rod. [is volume is are}. and it encloses charge pnrzL. Because the charge distribution is long, no electric flux passes through the circular end caps; E-dA = EdA cos90.0°= D - The curved surface has E -dA = EdAcosO", and E must be the same strength everywhere over the curved surface. “(3- P2429 2 Gauss’s law, §E-dA =1, becomes 13 IdA: pfl'f L. Eo em! 5'0 Surface Now the lateral surface area of the cylinder is 2:1: rL : 2 13(25 r]L = M r L , Thus, E =‘ 2:0 radially away frorn the cylinder axis \. Ea fit“! an " - ...
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