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hmmm
Feb2-05, 12:41 AM
A uniformly charged wire with a charge density of 4 microCoulombs/meter lies on the x-axis between x=1m and x=3m. What is the y-component of the corresponding electric field at y=3m on the y-axis?

I'm not really sure where to go with this. I want to treat the rod as an infinite number of point charges but I'm not sure how to calculate (y-component of) the electric field caused by each of these points.

epsilon infinity
Feb2-05, 12:52 AM
suppose there is a point charge on the point y=3 , find the electric fireld there, actually the electric field there for suppose due to the charge on the X=1 m can be broken up into two components one along -x and another along +y axis..they can be computed separately by integrating ..in this case i think you only need to compute for the y aixs one...for the integration take elemental lengths dx for the wire...

glad to be of help,
Arpan Roy
royarpan@hotmail.com

HallsofIvy
Feb2-05, 06:56 AM
Note that, for each "dx" on one side of the point, there is a corresponding "dx" the same distance on the other side. The horizontal components of force of those will cancel but the vertical components will add.

MathStudent
Feb2-05, 05:09 PM
Since this problem is not symmetrical, the horizontal components of the \vec{E} do not cancel. The easiest approach is probably to calculate the vertical ( \vec{E_y} ) component seperately.

Draw a diagram of the situation with the given axis, and choose an arbitrary piece of charge dq of the wire.

Come up with an equation for the corresponding electric field \vec{dE} due to dq at the point (0,3).

Figure a way to represent dq in terms of dx so you can integrate with respect to x.
(hint: it involves the linear charge density )

Break the equation into the vertical component of \vec{E} and integrate with respect to x
(hint: \vec{dE_y} = \vec{dE} sin \theta where theta is the angle between a line parallel to the x axis and \vec{E} )
(another hint: you will have to come up with an equation for sin \theta in terms of x so you can integrate )

good luck :smile:
-MS