session.masteringphysics.com Ch 21 HW Problem 21.87 22 of 25 Pa net!= 1.60×10-14

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session.masteringphysics.com Ch 21 HW Problem 21.87 22 of 25 Pa net!= 1.60×10-14 N Previous Answers Two 1.20m nonconducting wires meet at a right angle One segment carries 4 00 C of charge distributed uniformly along its length, and the other carries - 4.00 Correct C distributed uniformly along it, as shown in the figure (Figure 1). Part 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) Figure 1 of 1 1.20 m countercockwise from the + y- axis Submit Previous Answers Request Answer .20 m X Incorrect: Try Again; 3 attempts remaining Provide Feedback Next >

Explanation / Answer

Since point P is symmetrical and equidistant from both the nonconducting wires so field strength due to both will be equal.

Field due to positively charged wire will be vertically downward and field due to negative l charged wire will be leftward.

So net field at P will be at an angle of 225° from +x direction.

So electron will experience force in opposite direction since it's negatively charged particle.

So direction of force on electron will be 45° CCW from +x direction.

But in the question it has been asked from CCW from +y direction so it will be 315°.

So the correct answer will be 315°

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