What is the Method for Expressing Distances to Multiple Charges on the z-axis?

[tony873004]

I know I can express the distance from a point on the z-axis from the charge 2q as sqr(2a²+z²).

But I don't know how to express the distances to the other charges. I guess I could start by saying:

Let A be the point where charge q sits, B the point where charge 2q sits. Then the distance from a point on the z-axis to charge q would be sqrt((|AB|-a) ²+a²+z²)

and then do the same for the other charges. But this seems like a lot of work. I wouldn't imagine I'm supposed to start making up names for the points. Am I barking up the right tree with this method?

 

[neutrino]

Aren't they all just the same? The four corners of the square are sqrt{2}a units away from its centre.

For example, the location of -3q is (a,-a). So the distance to the origin from that point would be

\sqrt{\left(a-0\right)^2 + \left(-a-0\right)^2} = \sqrt{2a^2}

 

[Dick]

So what's AB? It looks to me from the picture like all of the charges are supposed to be the same distance from a point on the z axis. I.e. the origin O is the center of the square.

 

[tony873004]

That's what I thought, but I measured it. It's clearly off-center by a few millimeters. It even looks off-center which is why I double checked. This problem has 3 diamonds next to it which means it's supposed to be a butt-kicker. If the z-axis were in the middle, then it would be easy.

 

[Dick]

That's what I thought, but I measured it. It's clearly off-center by a few millimeters. It even looks off-center which is why I double checked. This problem has 3 diamonds next to it which means it's supposed to be a butt-kicker. If the z-axis were in the middle, then it would be easy.

If that were true, wouldn't they label the off-center distance?

 

[tony873004]

It's the lack of labeling that has me concerned, and makes me think this is a trick question. If it were centered, I would expect them to throw in a few more <-a-> 's to remove all ambiguity. Even if I measured it, and it were perfect, we shouldn't be expected to measure figures in the textbook. I'd still be nervous about assuming that it were centered.

 

[Saladsamurai]

Just a note tony, you may want try it and see what you get if they were the same distances and see if it matches the solution. I don't know if that's a Walker physics text, but mine is. . and sometimes I wonder about their "diamonds of difficulty" system.

Sometimes the "no diamond" problems make me want to eat razor blades while a three diamond is a walk in the park.

Caesy

EDIT: Sorry, just noticed it's an even-numbered problem=no solution to check. . . and I spelled my name wrong!

 

[tony873004]

. . . and I spelled my name wrong!

and I'm taking Physics advice from you?!

I'm not using Walker. I'm using "Physics, the Nature of Things" by Lea. The 3-diamond questions do tend to be tough in Lea. I like Walker better. Lea over-explains stuff, and Walker gets straight to the point. But Lea tends to be meticulous, which is why I'm doubting that the z-axis is centered in the illustration. It looks off-center, it measures off-center, and it's not labelled properly, missing some <-a-> s that a properly-labeled, z-axis in the middle drawing should have.

 

[Littlepig]

That's what I thought, but I measured it. It's clearly off-center by a few millimeters. It even looks off-center which is why I double checked. This problem has 3 diamonds next to it which means it's supposed to be a butt-kicker. If the z-axis were in the middle, then it would be easy.

well, I'm not sure, but if it isn't in the center, and all data is given in figure, how is suppose to you find out the distance of q, 6q and -3q to the center?? use center square.

 

[tony873004]

You were all correct. We were told to assume the z-axis is centered.