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Unformatted text preview: 8. The individual magnitudes E1 and E2 are ﬁgured from Eq. 23-3, where the absolute value signs for q are unnecessary since these charges are both positive. Whether we add the magnitudes or subtract them depends on if E1 is in the same, or opposite, direction as E2 . At points to the left of q1 (along the −x axis) both ﬁelds point leftward, and at points right of q2 (at x > d) both ﬁelds point rightward; in these regions the magnitude of the net ﬁeld is the sum E1 + E2 . In the region between the charges (0 < x < d) E1 points rightward and E2 points leftward, so the net ﬁeld in this range is Enet = E1 − E2 in the ˆ direction. Summarizing, we have i q q − 1 − (d+|2 |)2 2 x 1 q1 x q2 =ˆ i x2 − (d−x)2 4πε0 q1 2 + q2 2 x (x−d) for x < 0 for 0 < x < d for d < x Enet . We note that these can be written as a single expression applying to all three regions: Enet = 1 4πε0 q2 (x − d) q1 x + |x|3 |x − d|3 ˆ. i For −0.09 ≤ x ≤ 0.20 m with d = 0.10 m and charge values as speciﬁed in the problem, we ﬁnd
2e+07 E x –2e+07 ...
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