Problem 24.88 An infinitely long positively charged wire that carries a uniform

Question

Problem 24.88

An infinitely long positively charged wire that carries a uniform linear charge density passes through the origin of an xy coordinate system and lies on the y axis. At the position x=d on the xaxis, the electric field created by this wire has magnitude E and is directed to the right along the positive x axis. A charged particle is then placed on the x axis at x=2d. At x=d on the x axis, the electric field created by this particle has magnitude 2E and is directed to the left along the x axis.

Part A

At what positions on the x axis is the vector sum of the electric fields zero?

SubmitMy AnswersGive Up

Correct

Part B

What is the magnitude of the electric field produced by the wire at the locations you identified in the previous part?

Express your answer in terms of the variables d, , and the Coulomb's law constant k.

Explanation / Answer

at x = d

electric field due to long wire = E = 2*k*lambda/d......(1)

electric field due to point charge = 2E = k*q/d^2....(2)

1/2

1/2 = 2*lambda*d/q

lambda = q/4d

at point x the vector sum = 0

E1 = E2

2*k*lambda/x = k*q/(2d-x)^2

2*k*q/4dx = k*q/(2d-x)^2

1/2dx = 1/(2d-x)^2)

4d^2 + x^2 - 4dx = 2dx

4d^2 + x^2 - 6dx = 0

x = (3 + sqrt5)d

x = (3-sqrt5)d   <<<<<--------answer

++++++++++++++++++++++

partb

E' = 2*k*lambda/x

E' = 2*k*lambda/(3-sqrt5)*d

E' = E/(3-sqrt5)

E' = E/0.77 <<<<------aswer

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.