Solution for 1 D PHYS206 Lan2017spring x Secure I h morgan blackboard.co Spring

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Solution for 1

D PHYS206 Lan2017spring x Secure I h morgan blackboard.co Spring Assignment Week 2R. bdav/pid-1950627-dt content-rid-5851986 1/cou rses/16738.201730/PHYS206 Apps Harm Radio For Du D Diamond Ante AGeneral formation Dx Butternut HF6v 6-Ban Welcome to Security Video Groups G Convert And Resize A D Watch CNN live ICNI Name of Student: 1. (30 points) (Problems 43 in textbook Chapter 23 A continuous line of charge lies along the z axis, extending from z +zo to positive infinity. The line carries positive charge with a uniform linear charge density Ao. What are (a) the maga ide and (b) the direction of the electric fiekd at the origin? 2. (30 points) (Problems in textbook Chapter 23) A thin rod of length l and form charger per unit length A lies along the ar axis as shown bekw. (a) Show that the electric field at P. a distance d from the rod along its perpendicular bisector, has no r component and is given by E 2ke Asin 6o/d. (b) What if? Using your result to part (a), show that the field of a rod of infinite length is E 2k. A d. 3:24 AM 2/6/2017

Explanation / Answer

1.

The field dE at the origin due to an element of charge dq located at some x is;
dE = kdq/x^2

You now add up the fields due to all the dq's in the line. To do this you first have to express dq in terms of the density D (Ican't do lamda) and dx as;
dq = Ddx

dE = kDdx/x^2
Now integrate from x(o) to infinity, remember k & D are constant;

E = kDINTEGRAL[dx/x^2]
= -kD[1/(infinity) - 1/x(o)]
= kD/x(o)

The direction of E is to the left, along -x direction, since fields due to +charge always points away from the charge.

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