Resolução - Halliday - Volume 3 - Eletricidade E Magnetismo - P23 008

Transcript

8. The individual magnitudes
E and

E2
are figured from Eq. 23-3, where the absolute value signs for 1 q are unnecessary since these charges are both positive. Whether we add the magnitudes or subtract
2 . At points to the left of q1 (along the
1 is in the same, or opposite, direction as E them depends on if E −x axis) both fields point leftward, and at points right
of q
2 (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




2


1 points rightward and E
1

−

E
2 points leftward, so the net field in this range is E
net =

E (0 < x < d) E in the ˆi direction. Summarizing, we have
net E  q q2 1   − x2 − (d+|x|)2 1 q q 1 2 = ˆi x2 − (d−x)2 4πε0   q12 + q2 2 x (x−d) for x < 0 for 0 < x < d for d < x . We note that these can be written as a single expression applying to all three regions:   1 q2 (x − d) ˆ q1 x
i. + Enet = 4πε0 |x|3 |x − d|3 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