Adding electric field vectorially

[shemer77]

Homework Statement

I just wanted to make sure I am doing this right, my problem is that I think the electric field in the x direction should be the same for each one because the r_ix is the same for each point. Or am I just thinking too hard?
http://dl.dropbox.com/u/265697/IMG.jpg

Homework Equations

The Attempt at a Solution

 

[dynamicsolo]

Homework Statement

I just wanted to make sure I am doing this right, my problem is that I think the electric field in the x direction should be the same for each one because the r_ix is the same for each point. Or am I just thinking too hard?

You are correct that r_ix for the distance from the point at which the field is being measured to each of the cells is just the same. But the x-component of the electric field from each cell, $\Delta E_{ix}$ , is going to differ with each cell for two reasons: the total distance to the point from each cell is getting larger, so the strength of the electric field from each cell is getting successively smaller; and the angle that the direction from each cell makes with the x-axis is opening up further for each successive cell, so $\cos \theta$ is also getting smaller for each cell down the line. So $\Delta E_{ix}$ should get smaller for each cell down the list.

 

[shemer77]

thanks!

 

[shemer77]

Hey one more quick question for the net field box under $\Delta E_{ix}$ and $\Delta E_{iy}$ I just add up all the values right? and then what should I put in for the 2 boxes of the field magnitude and direction?

 

[dynamicsolo]

Hey one more quick question for the net field box under $\Delta E_{ix}$ and $\Delta E_{iy}$ I just add up all the values right? and then what should I put in for the 2 boxes of the field magnitude and direction?

Those are perpendicular components of the field $\Delta E_{i}$ from each cell, so they must be added "in quadrature", that is, according to the Pythagorean Theorem, since the field represents the hypotenuse of a right triangle with the two components being the triangle's legs. So $\Delta E_{i} = \sqrt{(\Delta E_{ix})^{2} + (\Delta E_{iy})^{2} }$.

EDIT: I had to go back and look at your table again. Yes, you would first add up the columns for $\Delta E_{ix}$ and $\Delta E_{iy}$ to get the totals of each component. Then you would combine the total x-component and y-component of the field in the way I described, but with $\Delta E = \sqrt{(\Delta E_{x})^{2} + (\Delta E_{y})^{2} }$. (I didn't remember what your table entries looked like...)

EDIT: For the direction of the total field, you need $\tan \theta$, which is equal to $\frac{\Delta E_{y}}{\Delta E_{x}}$ with the values taken from your sums . You would then calculate the inverse tangent (tan^-1) of that value to find $\theta$, which is the angle the field direction makes with the x-axis.

 

[shemer77]

thanks for your help! Really appreciate it, i got it and understand it now!