roblem 21.99 Two 1.20m nonconducting wires meet at a right angle. One segment ca
ID: 1592901 • Letter: R
Question
roblem 21.99 Two 1.20m nonconducting wires meet at a right angle. One segment carries 4.00 AuC of charge distributed uniformly along its length, and the other carries 4.00 C distributed uniformly along it, as shown in the figure (Figure 1) Figure 1 of 1 1.20 m P 1.20 m Part A Find the magnitude of the electric field these wires produce at point P, which is 60.0 cm from each wire. IV N/C Submit My Answers Give U Incorrect; Try Again Part B Find the direction of the electric field these wires produce at point P, which is 60.0 cm from each wire.(Suppose that the y-a countercockwise from the +y-axis Submit My Answers Give U Incorrect, Try Again Part C If an electron is released at P, what is the magnitude of the net force that these wires exert on it? IV F netExplanation / Answer
For each wire,
E = (k/z)(b/(z²+b²) + a/(z²+a²)
where k = 8.99e9N·m²/C²
and = ±4.0µC/1.2m (depending on the wire)
and z = 0.60 m
and a = b = L/2 = 0.60 m
Plugging in these values, I get
E = ± 61 803 N/C depending on the wire.
(b) The direction of the field at P is down (due to the top wire) and left (due to the left wire) in equal measure, so the direction is 45º below the -x axis, or 225º measured ccw from the +x axis.
(a) magnitude |E| = 61803N/C * 2 = 87 403 N/C
(c) mag |F| = q*|E| = 1.40e-14 N
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