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 .
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This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.
 Fall '08
 SPRUNGER
 Physics

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