Finding where electric field is zero with opposite charges

[flyingpig]

E = kq/(x-1)² works for either charge when to the right of x = 1.
That is, a positive value of q results in a positive E, meaning it points to the right. A negative value of q gives a negative E. But it goes wrong when x < 1.

Can yuo tell me why it cannot be on the right side of q2 of my problem? Because so far the construction is set to find the left, initially we never know.

 

[Delphi51]

Can yuo tell me why it cannot be on the right side of q2 of my problem?

The E1 and E2 due to q1 and q2 cannot be equal to the right of q2 because there we are closer to the larger charge; E2 will always be larger than E1.

 

[flyingpig]

The E1 and E2 due to q1 and q2 cannot be equal to the right of q2 because there we are closer to the larger charge; E2 will always be larger than E1.

What happens if the magnitudes are the same? Then how we could decide? Or will the quadratic really work out nicely?

Are there situations when I should use (d + x)?

 

[flyingpig]

Also, I noticed in the picture, the way you present the E field vector is almost analogous to the electrostatic force vector.

 

[Delphi51]

If the magnitudes are the same and in opposite directions, they cancel out and the total E is zero (as in the left side of my diagram). The quadratic equation finds the exact x value where this occurs.
No, I would never use d + x. The distance from q2 to position x is x-d. Or d-x if you don't care about the sign.

the way you present the E field vector is almost analogous to the electrostatic force vector

Yes, highly analogous. E is the same as the force vector on a charge of 1C.

 

[flyingpig]

If the magnitudes are the same and in opposite directions, they cancel out and the total E is zero (as in the left side of my diagram). The quadratic equation finds the exact x value where this occurs.
No, I would never use d + x. The distance from q2 to position x is x-d. Or d-x if you don't care about the sign.

No, I mean if q2 = +2.5uC and q1 = -2.5uC, I don't mean q2 = q1 in everyway

 

[Delphi51]

I don't mean q2 = q1 in everyway

Me either. I said E1 and E2 could cancel at a particular spot so E total is zero - the question originally posted here.

 

[flyingpig]

Me either. I said E1 and E2 could cancel at a particular spot so E total is zero - the question originally posted here.

[PLAIN]http://img833.imageshack.us/img833/6673/51209057.jpg

Should I use d-x again, but except I will only take positive roots?

$$\frac{q_{1}}{(d-x)^2}=\frac{q_{2}}{x^2}$$

Should it be x² or actually d²?

 

[Delphi51]

In view of our experience with the earlier question, better figure out which of the three regions the answer lies in before writing an equation!
Note that this one is more complicated because neither charge is at the origin.

 

[flyingpig]

In view of our experience with the earlier question, better figure out which of the three regions the answer lies in before writing an equation!
Note that this one is more complicated because neither charge is at the origin.

It has to be in between right?

 

[Delphi51]

Definitely.

 

[flyingpig]

So shouldn't this be correct?$$\frac{q_{1}}{(d-x)^2}=\frac{q_{2}}{x^2}$$

Should it be x² or actually d²?

 

[Delphi51]

That should work if you change the coordinate system so the origin is at Q2 and Q1 is at 10 meters. Then x should work out to some number between 0 and 10.

If you use the given coordinate system, you would use
Q1/(6-y)² = Q2/(y+4)²
and expect an answer for y between -4 and 6.

 

[flyingpig]

That should work if you change the coordinate system so the origin is at Q2 and Q1 is at 10 meters. Then x should work out to some number between 0 and 10.

If you use the given coordinate system, you would use
Q1/(6-y)² = Q2/(y+4)²
and expect an answer for y between -4 and 6.

I know it doesn't matter in this case, but I only need to plug in for the magnitudes of the charges when I get my quadratic right?

Also, even though I know it is in -4 and 6, why cannot it not be outside? I mean as they repel each other, the distance will not be 10 anymore right?

 

[flyingpig]

I got y = 0.1189m...

 

[Delphi51]

That answer is correct!
It can't be above Q1 because in that region the electric field due to both negative charges is down. Two downs can't add up to zero. Similarly, below Q2, both electric fields are up.

 

[flyingpig]

That answer is correct!
It can't be above Q1 because in that region the electric field due to both negative charges is down. Two downs can't add up to zero. Similarly, below Q2, both electric fields are up.

But it can't be 0.1189m

What if I square root both sides? Then I am forced to take the magnitudes only and which I will get 0.85m instead

 

[Delphi51]

Oops, terribly sorry! I worked it out using Q1/(6-y)² = Q2/(y+4)² and carefully checked the answer to make sure it was correct. But I never compared your answer with mine. Yours is different. Maybe show your work so we can compare.

 

[flyingpig]

Don't leave me! I got an exam in two days and I still got to get my understanding for Gauss's Law

Q1/(6-y)² = Q2/(y+4)²

Then square root both sides

√Q1/(6-y) = √Q2/ (y+4)

Cross multiply

√Q1(y+4) = √Q2(6-y)

√Q1y + 4√Q1 = 6√Q2 - y√Q2

y(√Q1 +√Q2) + 4√Q1 - 6√Q2 = 0

y(3 + √8) + 12 - 6√8 = 0

y = (12 - 6√8)/ (3 +√8) = -0.853m
So plug in the numbers and since we got only the square root, we must use the magnitudes and this is what happens even if we end up with the quadratic, we have to use the magnitudes

 

[flyingpig]

But if I take the quadratic

y² +168y - 144 = 0

I get y = 0.853m where I used the magnitudes of the charges

 

[ehild]

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 q_i and its position vector r_i. The electric field at a place with position vector r is

E=sum{kq_i(r-r_i)/|r-r_i|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

 

[flyingpig]

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...

 

[flyingpig]

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...

 

[ehild]

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

 

[flyingpig]

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?

 

[flyingpig]

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)

 

[flyingpig]

http://answers.yahoo.com/question/index?qid=20110108180433AAhvR4G

This guy, the way he did it, it seems easier, but he made a mistake at the end. He set it up the way "I would"

 

[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?

Read my post more carefully.

ehild

 

[Delphi51]

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.

 

[flyingpig]

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...