I am unhappy about the answer to this problem

[flyingpig]

Homework Statement

[PLAIN]http://img339.imageshack.us/img339/2158/71254129.jpg

In the figure above, determine the point (other than infinity) at which the electric field is zero. Given q₁ = -2.50\muC and q₂ = 6.00\muC

Solution: 1.82m to the left of q₁

The Attempt at a Solution

So I set it up as

\frac{q_{1}}{x^2} = \frac{q_{2}}{(d-x)^2} where is x is the point I am looking for

After some simplification

q_{1}d^2 - 2dq_{1}x + x^2(q_{1} - q{2}) = 0

Here is what hit me, if I was only going to concern about the magnitude, I would get x = -1.82m and 0.39m

First of all, I am not even sure why it is -1.82m and not 0.39. Isn't -1.82m outside the range of d = 1.00m?

Secondly, if I do not concern with the magnitude (using -2.50µC instead of 2.50µC), I couldn't even solve the quadratic.

So why must I use positive 2.50µC? And why is it -1.82m and not 0.39m?

Thanks!

 

[Pengwuino]

Recall that an electric field will give rise to a force on this charged particle you plan on introducing at some position 'x'. So you've sent to find where the field and thus, the force would vanish. If it is indeed 0.39m to the right of q1... how would that physically be possible?

If you introduced a test charge that is negative, it would receive a pull from q2 and a push from q1 - obviously not a point of vanishing electric field. If the test charge is positive, the opposite occurs - the charge receives a push from q2 and a pull from q1. This point can't be a point with no electric field.

Now consider the -1.83m position (or really, just look at the left side of q1) and run through the same arguments and hopefully you can see why the charge has to be on the left side of q1.

The charge does not have to be placed in between the charges. Infact, if q1 and q2 are of opposite charge, there exists no point between the charges with 0 electric field by the argument above. If q1 and q2 are of the same charge, the only two points of 0 electric field are in between q1 and q2.

 

[flyingpig]

Recall that an electric field will give rise to a force on this charged particle you plan on introducing at some position 'x'. So you've sent to find where the field and thus, the force would vanish. If it is indeed 0.39m to the right of q1... how would that physically be possible?

Of course it could!

[PLAIN]http://img207.imageshack.us/img207/9290/15058769.jpg

If you introduced a test charge that is negative, it would receive a pull from q2 and a push from q1 - obviously not a point of vanishing electric field. If the test charge is positive, the opposite occurs - the charge receives a push from q2 and a pull from q1. This point can't be a point with no electric field.

Now consider the -1.83m position (or really, just look at the left side of q1) and run through the same arguments and hopefully you can see why the charge has to be on the left side of q1.

If it is to left for real

1) A positive test charge would be attracted to the negative charge q₁, but the charge q₂ is greater, but also further and hence would make a "0" net charge

2) A negative test charge would be repelled by q₁, at the same time attracted to q₂

The charge does not have to be placed in between the charges. Infact, if q1 and q2 are of opposite charge, there exists no point between the charges with 0 electric field by the argument above. If q1 and q2 are of the same charge, the only two points of 0 electric field are in between q1 and q2.

But, just going back to my quadratic, why does only magnitudes of the charge work?

 

[gneill]

1) A positive test charge would be attracted to the negative charge q₁, but the charge q₂ is greater, but also further and hence would make a "0" net charge

But q₂ is a positive charge. It would repel your test charge, pushing it towards q₁ which is also attracting it. It would not be a "0" net field (not charge!). It would not be a point of no force for the positive test charge.

2) A negative test charge would be repelled by q₁, at the same time attracted to q₂

Right. So not a zero force point for the negative test charge either.

But, just going back to my quadratic, why does only magnitudes of the charge work?

You're already encapsulating the geometry of the situation into the equation. In particular, you're taking into account the directions of the fields caused by each charge in relation to their placement when you write the equation the way you did.

If you wanted to make your life more difficult you could write an equation that would make no assumptions about polarity of charge and the resulting force directions as you move from one side of a charge to another. This would necessarily be a vector equation. Then you could spend a quiet hour or two figuring out how to work vectors into a quadratic equation...

 

[flyingpig]

Could you dumb that last bit for me?

 

[Liquidxlax]

remember the electric field of a positive charge flows outwards and negative flows inwards. so you should have to be on the right of 2 or on the left of 1 to cancel out the e field

q1 and q2 are of different magnitudes and different sign

 

[Pengwuino]

Don't confuse force with charge.

If the charged particle is placed between Q1 and Q2 being pushed from Q1 and pulled from Q2 (thus, a negative test charge), the forces are adding and the result is a particle accelerating to the right. With a positive test charge, it would be pulled to the left and pushed from the right.

 

[Liquidxlax]

Don't confuse force with charge.

If the charged particle is placed between Q1 and Q2 being pushed from Q1 and pulled from Q2 (thus, a negative test charge), the forces are adding and the result is a particle accelerating to the right. With a positive test charge, it would be pulled to the left and pushed from the right.

my bad, its 2am... ignore my post then

 

[Delphi51]

In your first equation in the first post, the (d - x) assumes the point is between the charges. Use (d + x) to get the answer.

 

[SammyS]

Homework Statement

[PLAIN]http://img339.imageshack.us/img339/2158/71254129.jpg

In the figure above, determine the point (other than infinity) at which the electric field is zero. Given q₁ = -2.50\muC and q₂ = 6.00\muC

Solution: 1.82m to the left of q₁

The Attempt at a Solution

So I set it up as

\frac{q_{1}}{x^2} = \frac{q_{2}}{(d-x)^2} where is x is the point I am looking for

After some simplification

q_{1}d^2 - 2dq_{1}x + x^2(q_{1} - q_{2}) = 0

Here is what hit me, if I was only going to concern about the magnitude, I would get x = -1.82m and 0.39m

First of all, I am not even sure why it is -1.82m and not 0.39. Isn't -1.82m outside the range of d = 1.00m?

Secondly, if I do not concern with the magnitude (using -2.50µC instead of 2.50µC), I couldn't even solve the quadratic.

So why must I use positive 2.50µC? And why is it -1.82m and not 0.39m?

Thanks!

Hi flying pig.

To the left of q₁, which is at x = 0, the electric field due to q₁ is:

E_1=-k\frac{q_1}{x^2}\,.

To the right of q₁, the electric field due to q₁ is:

E_1=k\frac{q_1}{x^2}\,.


Similarly, to the left of q₂, which is at x = d, the electric field due to q₂ is:

E_2=-k\frac{q_2}{\left(x-d\right)^2}\,.

To the right of q₂, the electric field due to q₂ is:

E_2=k\frac{q_2}{\left(x-d\right)^2}\,.

The vector nature of the electric field in the above, is indicated by the signs. The effect of the signs of the charges on the electric field is built in.

The problem asks you to find a finite value of x at which the electric field is zero. That is:

\text{Solve }E_1+E_2=0

To the left of both charges and to the right of both charges:

k\frac{q_1}{x^2}+k\frac{q_2}{\left(x-d\right)^2}=0\quad\to\quad \frac{q_1}{x^2}=-\frac{q_2}{\left(x-d\right)^2}\,.
The RHS has the opposite sign of the RHS of your equation. However, IF you only consider magnitude (Why would you do that?), this is equivalent to your equation. Of course, the only way for this equation to be true, is for q₁×q₂<0.

Between the charges:

k\frac{q_1}{x^2}-k\frac{q_2}{\left(x-d\right)^2}=0\,.
This only has a solution if q₁×q₂>0.

 

[flyingpig]

It's more confusing if you use x - d even though the end result is the same because of the square

But why would I use (d+x)? I am really confused.

 

[Delphi51]

Sorry! I was thinking of x as a positive number. Your d-x is great; x just comes out to a negative number to indicate a point left of the origin.

 

[flyingpig]

I read Sammy's post over and over again which seems to be answering my question, but it just doesn't hit my head yet.

 

[flyingpig]

Wait, I found this link

http://www.jca.umbc.edu/~george/html/courses/phys112/2004fall/homework/04fall_phys112_hk1.pdf, go to question 27.

It's answering about the "magnitude" part, but really isn't telling me WHHY

I got the idea that it is a negative and positive attraction and it therefore would never make sense if it was x = 0.39 (why DOESN'T it make sense though? The math says it can)

 

[Pengwuino]

I got the idea that it is a negative and positive attraction and it therefore would never make sense if it was x = 0.39 (why DOESN'T it make sense though? The math says it can)

The math doesn't care about the physics. One of the skills you must build up is determining what solutions make sense and which do not. This is kind one of those cases.

Also, I didn't notice it but you asked why you must use the magnitudes of the charges. Well, for one, you equate two equations with strictly positive denominators. If one side has a charge that is negative and one that is positive, there's no way they can be equal simply because one is always positive and one is always negative.

Compute the forces at the point near the middle. They will be equal but in the same direction so it's not a solution.

 

[flyingpig]

The math doesn't care about the physics. One of the skills you must build up is determining what solutions make sense and which do not. This is kind one of those cases.

Sammy made me just realize something last night.

The forces are attractive and hence x can and should never be x = 0.39m even though the math worked out because I am saying there exists a charge in the middle such that the E-field is 0. But what does x = 0.39m mean then?

So when the charges are repulsive, it would be right to use d - x, but otherwise, d + x?

Also, I didn't notice it but you asked why you must use the magnitudes of the charges. Well, for one, you equate two equations with strictly positive denominators. If one side has a charge that is negative and one that is positive, there's no way they can be equal simply because one is always positive and one is always negative.

That makes so much sense I didn't even see it, I will now repeat the fourth grade!

 

[Delphi51]

I played with this some more and am surprised at how complicated it is!
I started with q1/x² + q2/(d-x)² = 0 and it worked perfectly, using the q1 = -2.5 and q2 = 6 and producing x = -1.82. Also a positive x spurious answer.

BUT if q1 happened to be a positive number so the solution is an x between 0 and 1, it would NOT work because both the q1/x² and the q2/(d-x)² are positive as if the E field from both charges is to the right. There should be a solution between 0 and 1 but my equation cannot find it.

I cannot think of a way to write the E1 + E2 = 0 equation that works for solutions in all three regions of x. The physicist must determine the zone where the solution lies before writing the equation!

 

[flyingpig]

Alright, I set it up again, but I still can't solve it

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

-|q_{1}| + 2x|q_{1}|+ x^2(|q_{1}|- |q_{2}|) = 0

 

[flyingpig]

I played with this some more and am surprised at how complicated it is!
I started with q1/x² + q2/(d-x)² = 0 and it worked perfectly, using the q1 = -2.5 and q2 = 6 and producing x = -1.82. Also a positive x spurious answer.

BUT if q1 happened to be a positive number so the solution is an x between 0 and 1, it would NOT work because both the q1/x² and the q2/(d-x)² are positive as if the E field from both charges is to the right. There should be a solution between 0 and 1 but my equation cannot find it.

I cannot think of a way to write the E1 + E2 = 0 equation that works for solutions in all three regions of x. The physicist must determine the zone where the solution lies before writing the equation!

How did you use the equation and used negative signs of the charges? It only works when you use the magnitude and I think that is why I got it wrong. Because when I set them equal to each other, that already took care of the signs.

But why did you use (d-x)? It really should be d +x

[PLAIN]http://img89.imageshack.us/img89/3586/76371619.jpg[/quote]

[PLAIN]http://img440.imageshack.us/img440/1547/47371918.jpg[/quote]

 

[Delphi51]

My slightly different equation worked perfectly with the -2.5 and +6.
I was thinking E1 + E2 = 0 and E1 was negative while E2 was positive, so it worked out.

But it wouldn't work in a situation where both charges are positive. In that case there should be a solution with x between 0 and 1, but my equation fails - both terms are positive and can't add up to zero.

These equations do not take care of the signs for the E1 and E2 terms properly, due to the fact that squaring the distance wipes out its sign.

 

[flyingpig]

Why did you have d -x ?

 

[Delphi51]

(d-x) works fine. The answer is -1.82, so (d-x) = (1 - (-1.82)) = 2.82, which is the correct distance. There is nothing wrong with the (d-x). The problem lies with it being squared - that wipes out the sign so q/(d-x)² can't indicate the direction of the E field at different values of x. The same applies to the q/x² term.

 

[flyingpig]

(d-x) works fine. The answer is -1.82, so (d-x) = (1 - (-1.82)) = 2.82, which is the correct distance. There is nothing wrong with the (d-x). The problem lies with it being squared - that wipes out the sign so q/(d-x)² can't indicate the direction of the E field at different values of x. The same applies to the q/x² term.

What if (d +x) then?

 

[Delphi51]

(d+x) doesn't work at all. Say we have a problem where the final answer is going to be x = -1.82. Then (d+x) = (1 - 1.82) = -.82. This is not the correct distance between q2 and x. It should be 2.82. Of course it will work if you interpret (d+x) to be the sum of the absolute values of d and x. But only left of the origin.

 

[flyingpig]

(d+x) doesn't work at all. Say we have a problem where the final answer is going to be x = -1.82. Then (d+x) = (1 - 1.82) = -.82. This is not the correct distance between q2 and x. It should be 2.82. Of course it will work if you interpret (d+x) to be the sum of the absolute values of d and x. But only left of the origin.

But my geometry makes it look like it can work...

 

[Delphi51]

Another way to think of it. Say you have a positive charge at x = 1. At a location where x > 1, the E field is to the right and E = kq/(x-1)² handles the sign okay. But it doesn't work for x < 0 where the E field is to the left and should be negative. Is there any way to modify the formula to make it handle the sign of E? How about
E = kq/[(x-1)*|x-1|]
But how would you work with that in your quadratic equation?

 

[flyingpig]

Is it because we know that it must be to the left? Why can't it be the right

 

[flyingpig]

Another way to think of it. Say you have a positive charge at x = 1. At a location where x > 1, the E field is to the right and E = kq/(x-1)² handles the sign okay.

But what test charge (plus or minus?) are you using to get that formula?

But it doesn't work for x < 0 where the E field is to the left and should be negative. Is there any way to modify the formula to make it handle the sign of E? How about
E = kq/[(x-1)*|x-1|]
But how would you work with that in your quadratic equation?[/QUOTE]

 

[SammyS]

Sammy made me just realize something last night.

The forces are attractive and hence x can and should never be x = 0.39m even though the math worked out because I am saying there exists a charge in the middle such that the E-field is 0. But what does x = 0.39m mean then?

The answer of x = 0.39m is correct if q₁ and q₂ are like charges, i.e. if they have the same sign.

 

[Delphi51]

But what test charge (plus or minus?) are you using to get that formula?

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.

 

[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 Guass'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