Two 1.20 -in nonconducting wires meet at a right angle. One...

Two 1.20 -in nonconducting wires meet at a right angle. One segment carries $+2.50 \mu \mathrm{C}$ of charge distributed uniformly along its length, and the other carries $-2.50 \mu \mathrm{C}$ distributed uniformly along it, as shown in Fig. 21.50 (a) Find the magnitude and direction of the electric field these wires produce at point $P,$ which is 60.0 $\mathrm{cm}$ from each wire. (b) If an electron is released at $P$ , what are the magnitude and direction of the net force that these wires exert on it?

Transcript

00:01 The first item of this question we have to calculate the induced electric field at the point p.

00:06 Notice that the point p is located exactly in the middle of both wires.

00:12 So it's half of this wire and half of this wire.

00:16 This means that the electric field will follow some symmetry.

00:21 The red wire carries a positive charge, so it produces an electric field that points like this, always away from the wire.

00:30 On the other hand, the blue wire carries a negative charge, so it produces an electric field pointing towards it, like this and like this.

00:40 So the contribution of each wire for the induced electric field on the point p will be as follows.

00:46 The red wire induces something like this, let me call it e plus.

00:52 At the same time, the blue wire induces something like this.

00:56 Let me call this e minus.

00:59 Notice that both wires have the same charge and both are located at the same distance from the point p.

01:07 Therefore, e plus and e minus will have the same magnitude.

01:11 It means that the magnitude of e plus is equal to the magnitude of e minus.

01:17 The magnitude of these fields is given by 1 divided by 4 pi epsilon 0 times the absolute value of the charge divided by x square root of x squared plus a squared.

01:33 Then remember that 1 divided by 4 pi epsilon 0 is equals to 9 times 10 to the 9, newton's divided by culum squared times meters squared.

01:46 Then the field is given by 9 times 10 to the 9 times the absolute value of the charge, which is 2 .5 times 10 to minus 6 because of the micro.

02:01 And this is divided by x, which is 0 .6, times the square root of 0 .6 squared plus a squared, which is also 0 .6 squared.

02:14 And this results in approximately 44 ,194 .2 neutrons per column.

02:24 Remember that this is for each contribution, for the negative wire and for the positive wire.

02:32 So now we have to compute what is the resulting electric field.

02:36 It's not difficult to do so.

02:38 The resulting electric field will be something like this, which is composed by e minus plus e plus...

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