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A uniformly charged ring of radius 10.0 cm has a total charge of 58.0 uC Find the electric field on the axis of the ring at the following distances from the center of the ring. (Choose the x-axis to point along the axis of the ring.) (a) 1.00 cm (b) 5.00 cm (c) 30.0 cm i MN/C (d) 100 cnm
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Answer #1

First we establish a formula for the electric field due to the charged ring at a distance 'x' from the center of the ring on it's axis.Then we can substitute any value of 'x' and get the required electric field at any distance.

Given that -he thl charge on the d Ex so the total length of ing is L ItR d E -s 5.8 x Io en si -σ 9.23x10-son . Now Consider a very small length of the this smalt sing at the top f the ing as sheun in the Agur length be dl, Se the charge will be dad Now Cons: der a n he 3xSThis point aft ba at a Snall charpe.Terefove he electic eld wold be de- kd9 point .p at a distance 2、m -he center of the kd9 R2 +22 Ths electne field will have uoo ompones, one alongx-s a the ther kda -axis an d alon The 1-Component in be in the sane direc tion br al Such Sm-eal charpas on -veǐing, but they-amponents du e-otop kal n downusard direchon and hence get cancelled due to the lower haifof the ng Chages So in effect we have 1 the charge on the in a+ same distance from the Point P e and al caibuting to the x-comparent only. So 艹1 net elecㅢ거cfeld will be. The Mng uoi be inSo, we have our -frmula for elechic 솎eld at a distance z Nou lets calculate the anSwars. Ca) (9x15)(5.8 x 105)(10- oi) Icm El = 1.015 x 10 3 X10 )(5.8 x1。 (0.05 61 x 10 0-12 0.052) 1.3975 x10 =18.676 . MIN (c) χ3 =(9x10,)(5.8x10 s)(0.3) = 30 cm= 0.3 m ; E2 156.6 X103 3.16 x to 5 = 4.952 I MN 100 cm E4 = (9x109)(5.8x10 m ; 0.1

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