Two d = 1.40-m nonconducting wires meet at a right angle. One segment carries +4

Question

Two d = 1.40-m nonconducting wires meet at a right angle. One segment carries +4.20 mu C of charge distributed uniformly along its length, and the other carries -4.2 mu C distributed uniformly along it, as shown in the following figure. (a) Find the magnitude and direction of the electric field these wires produce at point P, which is 70.0 cm from each wire. (Assume the +x-axis goes to the right.) magnitude N/C direction degree counterclockwise from the +x direction (b) If an electron is released at P, what are the magnitude and direction of the net force that these wires exert on it? magnitude N direction degree counterclockwise from the +x direction

Explanation / Answer

magnitude of field at perpendicular bisector ,

E = k Q / ( (d/2) sqrt( (d/2)^2 + (d/2)^2))

E = (9 x 10^9 x 4.20 x 10^-6) / ( 0.70 sqrt(0.70^2 + 0.70^2))

E = 5.45 x 10^4 N/C

due to upper wire,

E1 = (5.45 x 10^4 N/C) (-j)

due to left wire,

E2 = (5.45 x 10^4 N/C) (-i )

magnitude = sqrt(5.45^2 + 5.45^2) x 10^4 = 7.71 x 10^4 N/C

direction = 180 + tan^-1(5.45 /5.45) = 225 deg

(B) F = q E

F = (-1.6 x 10^-19) (E)

magnitdue = 1.6 x 10^-19 x 7.71 x 10^4

= 1.234 x 10^-14 N

direction = 45 deg

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