I've got 9-11, but I'm having trouble finding the d's or the V's for 9-10. The change in V is throwing me off in the problem statement. Can anyone give me a hint? Thanks.

3/12/2007 5:58:58 PM

You know that V1 - V2 = 5V.

You know that V = Q / (4 * pi * epsilon * d).

The problem gives you a relationship between x and y, and the location of points 1 and 2. You can use this to find algebraic equations for the distances from the charge to each of the two points.

Plug your distance equation into the voltage equations and plug those into voltage difference equation.

3/12/2007 7:09:48 PM

All I did was say that 5V = (Field at Pt1) - (Field at Pt2)....

so I used a vector r0 to be the distance from point1 and 2.82-r0 to be the distance from point 2. Then you have an equasion that can be solved for r0 (the distance from point1).

I got the distance to be alittle over 1, which puts it slightly closer to point1 than point2...

Maybe im way off base.
Having trouble with the matlab though...

Im thinking you have to look at the field due to each of the 4 charges using superposition to sum the charges. What I don't know how to do is find a statement to look at each charge.

I'm thinking:

Find Value due to charge1
Find Value due to charge2
Find Value due to charge3
Find Value due to charge4

Sum

Plot

Then take a new value of X / Y and repeat


[Edited on March 12, 2007 at 7:15 PM. Reason : more]

3/12/2007 7:11:34 PM

Did anyone else get two answers for 9-10? I'm trying to decide which one is right.

3/12/2007 9:32:05 PM

Of your two answers, which one will produce a higher voltage at (10,10) and a lower voltage at (8,8)?

3/12/2007 10:19:18 PM

Both are closer to 8...

3/12/2007 10:29:52 PM

You should've ended up with a quadratic equation--one of the solutions is clearly closer to (10,10) and the other solution is clearly closer to (8,8).

You may want to try jbl4me's solution. Find the distance from the charge to each of the two points, and then find where the charge has to be to make the required distances work out. Essentially, plug d1 = r and d2 = r + 2.83 into the voltage equations and plug those into the difference equation. r is the distance from the charge to (10,10) and 2.83 is the distance between (10,10) and (8,8). Because you know the geometry of the points and you which point has the higher voltage, you can know that (8,8) is 2.83 further away from the charge than (10,10).

Hope that helps...I don't know what else to tell you.

3/13/2007 7:36:59 AM

Quote :

"so I used a vector r0 to be the distance from point1 and 2.82-r0 "

Shouldn't it be r0 + 2.82?

3/13/2007 7:53:29 AM

Not the way I solved it. In my approach, a r0 of 0 would mean that you are right on top of point one, and thus point 2 is 2.82 away from point one. Since we know the charge is on the line y=x, then as r0 increases from 0 (point1) then the distance from the particle to point 2 is 2.82-r0.

I could be wrong...

3/13/2007 1:53:42 PM

I actually thought that you just made a typo. Looking at it again, it looks like you assumed that the charge was between the two points. Is that right?

The homework is turned in, so let's compare answers.

I got (12.89,12.89).

Doing it jbl4me's way, I get (9.07,9.07)

3/13/2007 2:52:11 PM

Well, since the difference is positive, you know that the charge must be closer to p1 than p2. I guess that doesn't mean its in the center though...but its def closer to p1. So...my logic is flawed.

3/13/2007 2:57:32 PM

I thought it was obvious the charge was to the left of p1, since it has a higher voltage. I ended up getting the charge was located somewhere near 3.

3/13/2007 3:39:38 PM

Crap. I did my problem as V(10,10) - V(8,8) = 5, not the other way around.

3/13/2007 5:19:39 PM

^^ I was a little confused in your earlier post...now I know why. Nevertheless, I got around 5.

[Edited on March 13, 2007 at 7:54 PM. Reason : ]

3/13/2007 7:53:13 PM

Here is how I did it...

[ (((50/9)e-9)*9e9) / sqrt((x-8)^2 + (y-8)^2) ] - [ (((50/9)e-9)*9e9) / sqrt((x-10)^2 + (y-10)^2) ] = 5 V

y=x, so change all the y's to x's...and solve for x

3/13/2007 7:58:15 PM

That's what I did--I just chose the wrong root from the resulting quadratic equation.

3/14/2007 10:18:16 AM