Resolução - Halliday - Volume 3 - Eletricidade E Magnetismo - P23 016

Transcript

16. From the figure below it is clear that the net electric field at point P points in the −ˆj direction. Its magnitude is  

q d/2


Enet
= 2E1 sin θ = 2 k (d/2)2 + r2 (d/2)2 + r2 = k qd + r2 ]3/2 [(d/2)2 where we use k for 1/4πε0 for brevity. For r  d, we write [(d/2)2 + r2 ]3/2 ≈ r3 so the expression above reduces to

qd


Enet
≈ k 3 . r ˆ Since p = (qd)j,  net ≈ −k p . E r3 y +q • d/2 d/2 −q • P ............... ............. .... .......................... ............. ............. ... ............. .... ............. ... ............... ..... .............. ... ......................... ......... ...... .. ....... . ..... ... ...... .. . ....... .. . . . . .... .... . . ... ..... ......... .... ....... . .. ... ..  net E x