I am unhappy about the answer to this problem

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The discussion revolves around determining the point where the electric field is zero between two charges, q1 and q2, with values of -2.50µC and 6.00µC, respectively. The calculated position for zero electric field is 1.82m to the left of q1, leading to confusion about why a solution of 0.39m to the right of q1 is not valid. Participants clarify that a test charge placed at 0.39m would not experience a net zero electric field due to the opposing forces from the two charges. The importance of considering the signs and magnitudes of the charges in the equations is emphasized, as it affects the physical interpretation of the solutions. Ultimately, the discussion highlights the need for a thorough understanding of electric fields and forces when solving such problems.
  • #51
The electric field of a system of point charges is the sum of the contributions of all the individual charges. Let the magnitude of the i-th charge qi and its position vector ri. The electric field at a place with position vector r is

E=sum{kqi(r-ri)/|r-ri|3}.

(http://alumnus.caltech.edu/~muresan/projects/esfields/field.html)

Your need the field along the line connecting two charges, so the y and z coordinates of all position vectors are 0: you get an equation for the x coordinates. Put the origin at the place of q1.

q1 = -2.5 μC, x1=0
q2 = 6 μC, x2= 1 m.

You want to find the x position where the field is zero. Plugging in the given data into the equation above: -2.5 x/|x|^3+6(x-1)/|x-1|^3=0.

Take care of the absolute-value sign ||.
You know that |x|=x if x>0 and |x|=-x if x<0.
In the same way, |x-1|=x-1 if x>1 and |x-1|=-x+1 if x<1.
You have three possibilities: x<0 (on the left of q1), 0<x<1 (between q1 and q2), and x>1 (on the right of q2 ) Try to solve the equation for all cases.

ehild
 
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  • #52
ehlid, I noticed you used x - d (d = 1.00) instead of d - x...

interesting...

I think this is the third thread I have on this forum that deals with signs...
 
  • #53
ehlid, I noticed you used x - d (d = 1.00) instead of d - x...

interesting...

I think this is the third thread I have on this forum that deals with signs...
 
  • #54
You have tried quite long to get the correct result for this very simple problem. Three peoples have tried to explain the meaning of Coulomb's force. You should remember that the electric field strength is a vector quantity. You can not avoid directions and signs. Try to understand Coulomb's Law and how it has to be applied if there are more than one point charge.

ehild
 
  • #55
ehild said:
You have tried quite long to get the correct result for this very simple problem. Three peoples have tried to explain the meaning of Coulomb's force. You should remember that the electric field strength is a vector quantity. You can not avoid directions and signs. Try to understand Coulomb's Law and how it has to be applied if there are more than one point charge.

ehild

I also noticed you did

-2.5 x/|x|^3+6(x-1)/|x-1|^3 = 0

instead of

-2.5 (x-1)/|x-1|^3 +6x/|x|^3 = 0

Why?
 
  • #56
Wait, just going back to the new problem, the reason Delph set it up as

Q1/(6-y)² = Q2/(y+4)²

For (6-y), this is the units away from the origin and so is y + 4

The way I have been doing is that I am taking the absolute value of the distance doing

Q1/d² = Q2/(d-y)², but this could never work because I have charges below the origin (below the x-axis)
 
  • #58
flyingpig said:
I also noticed you did

-2.5 x/|x|^3+6(x-1)/|x-1|^3 = 0

instead of

-2.5 (x-1)/|x-1|^3 +6x/|x|^3 = 0

Why?

Read my post more carefully.

ehild
 
  • #59
You've got the answer! Looks like exactly the same work as I had:
Q1/(6-y)² = Q2/(y+4)²
-9/(6-y)² = -8/(y+4)²
9(y+4)² = 8(6-y)² after multiplying both sides by common denominator
y² + 168y - 144 = 0
y = 0.8528

I don't think it is a good idea to take the square root of both sides - you would have to be very careful about whether it is 6-y or 6+y if you did that! Nor do you need to take the absolute value of the distance when it is squared.

We used 9 and 8 instead of 9 x 10^-6 in the first step - okay since the 10^-6 on both sides would cancel out.

Ehild has a more sophisticated method that works without knowing in which zone the answer lies, but it is quite a bit more complex mathematically.
 
  • #60
Delphi51 said:
You've got the answer! Looks like exactly the same work as I had:
Q1/(6-y)² = Q2/(y+4)²
-9/(6-y)² = -8/(y+4)²
9(y+4)² = 8(6-y)² after multiplying both sides by common denominator
y² + 168y - 144 = 0
y = 0.8528

I don't think it is a good idea to take the square root of both sides - you would have to be very careful about whether it is 6-y or 6+y if you did that! Nor do you need to take the absolute value of the distance when it is squared.

We used 9 and 8 instead of 9 x 10^-6 in the first step - okay since the 10^-6 on both sides would cancel out.

In that link (yahoo) can you tell me what he meant by "Since r₁ is distance from Q1 to null point, then null point is y=0.46"

I know he just subtracted 4, but I don't understand his sentence well.

What happens if both charges are in the negatives??

Ehild has a more sophisticated method that works without knowing in which zone the answer lies, but it is quite a bit more complex mathematically.[/QUOTE]

And it's too much for my brain...
 
  • #61
Wait, something new even just occurred to me. The whole (6-y) and (y + 4) is just a change of distance or displacement

(6 - y), from some point y to a y = 6

(y + 4) = from some point y to a y = -4 = (-4 - y) = -(4+y)

But the -(4+y) has a negative sign, but if I square the whole thing (-(4+y))^2, I can get rid of it.

Can I think of it like that? Or is this just a mechanical coincidence?
 
  • #62
Since r₁ is distance from Q1 to null point, then null point is y=0.46
He calculates r₁ = 5.54, where r1 is the distance of the null point from Q1. Q1 is at y = 6. Hence the null point is 6 - 5.54 = 0.46.
This is slightly different way of doing the question, using distances from the charge to the null point and figuring out what the y coordinate is afterwards. You know, once you have the diagram drawn, you are free to use whatever coordinate system you like and the math will be simplest if you put the origin right where one of the charges is. Like the original post in this thread.

Can I think of it like that?
Yes. You can ignore the minus sign, use either (6-y) or (y-6) as long as you leave it squared - that's why I suggested you avoid taking the square root of both sides.
 
  • #63
Delphi51 said:
He calculates r₁ = 5.54, where r1 is the distance of the null point from Q1. Q1 is at y = 6. Hence the null point is 6 - 5.54 = 0.46.

What does null point mean?
Yes. You can ignore the minus sign, use either (6-y) or (y-6) as long as you leave it squared - that's why I suggested you avoid taking the square root of both sides.

Here try this

[PLAIN]http://img196.imageshack.us/img196/8453/38286539.png

If I use a symmetry argument, it is immediately known that it must be y = -15cm, but the math way to do it

Q1/(y+10)^2 = Q2/(y + 20)^2

Then y = 15cm, but not y = -15cm

If had set it up like

Q1/y^2 = Q2/(10 - y)^2, then I get y = 5cm instead
 
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  • #64
Ah wait no...never mind, 20y + 300 = 0 is y = -15, forgot the 0
 
  • #65
Good catch!
Note also the trouble you get into if you take the square root of both sides. Of course it does work out if you consider both the positive and negative square root.
The "null point" is the point where E = 0.
 
  • #66
I thought of this problem.

[PLAIN]http://img94.imageshack.us/img94/2664/33685990.jpg

Where Q1 = Q2 and Q1 is at (0, 10) and Q2 is at (10,10). from symmetry it must be where (5,10), but the math gives me (5,y)

Q1/(10-y)² = Q2/(10-y)²

Clearly, y can be anything, what does that mean?

My other question is, what if this was in polar coordinates? Does that make it easier if the charges sit in space instead of on axises?
 
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  • #67
Where Q1 = Q2 and Q1 is at (0, 10) and Q2 is at (10,10).

Q1/(x)² = Q2/(10-x)²
The y coodinate is y=10 for all points including the one where E = 0.

I wouldn't do it in polar coordinates - too messy.
 
  • #68
Delphi51 said:
Where Q1 = Q2 and Q1 is at (0, 10) and Q2 is at (10,10).

Q1/(x)² = Q2/(10-x)²
The y coodinate is y=10 for all points including the one where E = 0.

I wouldn't do it in polar coordinates - too messy.

But how do we show that it is y = 10?
 
  • #69
The method we are using only works for two charges on a line, with the E=0 point on the same line. In this last question, everything is on the line y = 10.

It would be interesting to try doing a question where the two charges are not on a horizontal or vertical line. The easiest approach would be to draw a straight line joining q1 and q2 and using their separation distance - essentially rotating the coordinate system so only one dimension matters. Otherwise, the equation would have both x's and y's. A second equation - requiring the null point to be on the line joining q1 and q2 - would be needed to solve for both the x and y coordinates of the null point.
 
  • #70
That's what I thought because

Q1/(10-y)² = Q2/(10-y)²

Will never work.
 
  • #71
That equation is just wrong. Note my correction in post #67.
 
  • #72
Delphi51 said:
That equation is just wrong. Note my correction in post #67.

I asked my instructor about it and he gave me a proof of a point where (I hope I am using this symbol right) P ∈ (0,10) in this x interval such that P must be at 5cm.
 
  • #73
Delphi51 said:
It would be interesting to try doing a question where the two charges are not on a horizontal or vertical line.


You have two point charges: Q1 at point 1 and Q2 at point P2. Find the electric field strength at point P. r1 is the vector that points from P1 to P and r2 points from P2 to P. r1 and r2 are the magnitudes. I will use bold for vectors. (TEX does not work for me.) The contribution of a point charge Q to the electric field strength at P is

E (P)= (kQ /r2) r(hat),

where r(hat) is the unit vector along the vector r. It is equal to r/r, so the previous formula can also be written as

E =(kQ /r3) r

The contributions of both charges add up and yield zero at P:

E(P)=( k Q1 /r13 )r1+ (k Q2 /r23) r2=0

E(P) is a linear combination of the vectors r1 and r2 and it is zero. That means that the vectors r1 and r2 are parallel, which can happen only when P lies on the straight line connecting P1 and P2.

Denoting the vector from P1 to P2 by R, R=r1-r2

r1=t1R and r2=t2R.

t1 and t2 being scalars. Denoting the magnitude of R by R (it is the distance between P1 and P2),

t1-t2=1

and the condition for E=0 at P can be written in the form

Q1 t1R/|t1R|3 + Q2 t2R /|t2R|3 =0

R/R3 factored out the equation simplifies to

Q1 t1/|t1|3 + Q2t2 /|t2|3 =0

or sign(t1) Q1 /t12 + sign(t2) Q2/t22 =0

with the condition that t1-t2=1.

If P1:(x1,y1) and P2:(x2,y2) the coordinates of P are

x=x1+t1(x2-x1),
y=y1+t1(y2-y1).

ehild
 

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