Two nonconducting wires of length L = 4.40 m meet at a right angle. One segment

Question

Two nonconducting wires of length L = 4.40 m meet at a right angle. One segment carries +4.10 mu C of charge distributed uniformly along its length, and the other carries -4.1 mu C distributed uniformly along it, as shown in the following figure. Find the magnitude of the electric field these wires produce at point P, which is equidistant from the midpoints of each wire. If an electron is released at P, what are the magnitude and direction of the net force that it experiences due to these wires?

Explanation / Answer

(a) For each wire,
E = (k/z)(b/(z²+b²) + a/(z²+a²)
where k = 8.99e9N·m²/C²
and = ±4.1 µC/4.4m (depending on the wire)
and z = 2.2m
and a = b = L/2 = 2.2 m

Plugging in these values
E = 5391 N/C due to each wire.

the elctric field due to the two wires are perpendicular to each other . so net field = sqrt (53912 +53912) =7624 N/c

(b) mag |F| = q*|E| = 1.6x10-19 (7624)=1.2x10-15 N

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.The electron moves in the direction opposite the field .so the direction of force is 45 degrees ccw from + x axis.

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