P23_016 - 16. From the figure below it is clear that the...

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Unformatted text preview: 16. From the figure below it is clear that the net electric field at point P points in the −ˆ direction. Its j magnitude is Enet = 2E1 sin θ = 2 k = k q (d/2)2 + r2 d/ 2 (d/2)2 + r2 qd + r 2 ] 3/ 2 [(d/2)2 where we use k for 1/4πε0 for brevity. For r reduces to d, we write [(d/2)2 + r2 ]3/2 ≈ r3 so the expression above Enet ≈ k qd . r3 Enet ≈ −k p . r3 Since p = (qd)ˆ j, y +q • d/2 d/2 −q • P . ..... .... ........... ....... . ............. ....... . ............. . . ....... . ..... . ....... ....... . . . .......... ....... .......... . ......... . . ...... .... ..... ...... .. . .. . ......... . ......... ... . . .. . . . ... .... ... .. . ... . .. .. ... .. .. ... ... . . ... . ..... ... . ... . . . ... . ... .. .. .. .... . . .. . .. .. ... ... . .. . . Enet x ...
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