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Unformatted text preview: 57. We note that the contributions to the ﬁeld from the pair of −2q charges exactly cancel, and we are left
with the (opposing) contributions from the 4q (at r = 2d) and −q (at r = d) charges. Therefore, using
k = 1/4πε0
|Enet | = k
−k 2 =0 .
The net ﬁeld at P vanishes completely. ...
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