I am unhappy about the answer to this problem

[flyingpig]

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?

 

[Delphi51]

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.

 

[flyingpig]

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

 

[flyingpig]

Ah wait no...never mind, 20y + 300 = 0 is y = -15, forgot the 0

 

[Delphi51]

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.

 

[flyingpig]

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?

 

[Delphi51]

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.

 

[flyingpig]

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?

 

[Delphi51]

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.

 

[flyingpig]

That's what I thought because

Q1/(10-y)² = Q2/(10-y)²

Will never work.

 

[Delphi51]

That equation is just wrong. Note my correction in post #67.

 

[flyingpig]

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.

 

[ehild]

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