Two 1.20m nonconducting wires meet at a right angle. One segment carries 3.50 C

Question

Two 1.20m nonconducting wires meet at a right angle. One segment carries 3.50 C of charge distributed uniformly along its length, and the other carries 3.50 C distributed uniformly along it, as shown in the figure (Figure 1) .

a.Find the magnitude of the electric field these wires produce at point P, which is 60.0 cm from each wire.

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-axis directed vertically.)

c.If an electron is released at P, what is the magnitude of the net force that these wires exert on it?

d.If an electron is released at P, what is the direction of the net force that these wires exert on it?(Suppose that the y-axis directed vertically.)

Explanation / Answer

E = (k*lambda/z)(b/sqrt(z^2+b^2) + a/sqrt(z^2+a^2)

k = 9 x 10^9

and lambda = +/- 3.50uC/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 = +/- 61803 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 45 degree below the -x axis, or 225 degree measured ccw from the +x axis.

(a) magnitude |E| = 61803N/C * sqrt(2) = 87403 N/C

(c) mag |F| = q*|E| =

q = 1.6 x 10^-19 C

= 1.40 x 10^-14 N

(d) an electron moves in the direction opposite the field, so

theta = 45 degree N of E

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