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# electrical forces v2

Posted on 2010-08-13
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HI

If A and B are oppositely charged objects and placed along an axis, why don't they move towards each other? Perhaps they do. There is an attractive force and nothing opposing it. If they did move towards each other, wouldn't the force between them then get bigger and bigger the closer they get? And then presumably they'd move together even faster.

thanks

thanks
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Question by:andieje
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Expert Comment

ID: 33429692
>> why don't they move towards each other?
Why would you think they will not move towards each other if there were no contraints or friction to hold them back?
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Expert Comment

ID: 33429868
o ->F                    F<- o
F = c/r^2
If objects have the same mass, then F = m a = c/r^2,
so a(t) = (c/m) / r(t)^2 for each object.

a1(t) = x1''(t) --> x1''(t) = (c/m) / r(t)^2
r = x2 - x1
x1''(t) = (c/m) / ( x2(t) - x1(t) )^2
x2''(t) = (c/m) / ( x2(t) - x1(t) )^2

You can run a computer simulation to find out their motion. If not the same masses, then need to make adjustments. If the two objects have different masses and are initially stationary, then if you draw a circle around the two objects, there are no external forces on them. So, I believe the center of mass of the two objects will not change as a function of time. This means that the collision will be at the center of mass (unless one object is so massive and large that the center of mass of the two objects is inside the boundary of the massive object.

Here is an Earth Moon collision in a spreadsheet to illustrate the point. It's almost correct.

http://www.experts-exchange.com/Other/Math_Science/Q_25280369.html
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Author Comment

ID: 33429895
So they do move towards each other?
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LVL 27

Expert Comment

ID: 33429914
Oppositely charged objects/particles attract, and the force of attraction is inversely proportional to the
square of the the distance between them.

Classical mechanics says the particles should move toward each other, and the rate of acceleration
should increase.

Quantum mechanical considerations keep atoms (electrons and protons combined with neutrons)
from collapsing.
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Expert Comment

ID: 33429919
yes, if oppositely charged, then they attract. If same charge, they repel. Attraction draws objects closer.
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Author Comment

ID: 33429957
I don't get the answer to this question (it was this question that made me think perhaps i misunderstood the direction of movement)

http://www.physicsclassroom.com/Class/estatics/u8l3d.cfm

Three charges are placed along the X-axis. Charge A is a +18 nC charge placed at the origin. Charge B is a -27 nC charge placed at the 60 cm location. Where along the axis (at what x-coordinate?) must positively charged C be placed in order to be at equilibrium?

At the most basic level i thought C must be placed at the right of B:
B will be in equilibrium when the force from A and from C are equal
A will be in equilibrium when the attractive force to B is equal to the repulsive force from C
C will be in equilibrium when the attractive force to B is equal to the repulsive force from A

but the answer says something totally different
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Expert Comment

ID: 33429962
>> If A and B are oppositely charged objects
These could be large charged objects as opposed to electrons and protons.

Just to be clear
>> "the force of attraction is inversely proportional to the square of the the distance between them."

This is equivalent to what I said:
>> F = m a = c/r^2

>> x1''(t) = (c/m) / ( x2(t) - x1(t) )^2
Since x2(t) - x1(t) is getting smaller, then x1''(t) and x2''(t) are increasing.
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Expert Comment

ID: 33430035
Maybe they are just talking about C being in equilibrium by having the forces of A and B being equal and opposite on C.
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Author Comment

ID: 33430058
They say C should be to the right of A. I don't see how that can be right in any context
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Author Comment

ID: 33430080
sorry i meant to the left of A. It says C is to the left of A
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LVL 32

Expert Comment

ID: 33430082
They say C should be to the left of A.
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Expert Comment

ID: 33430130
They are setting FAC = FBC to get the forces on C to balance out so that C does not move.

A is a weaker charge than B, so if C is to the left of A, then the stronger B charge will have the same force on C as A does even though the distance from C to B is greater than the distance from C to A.
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Author Comment

ID: 33430147
another thought: does that mean oppositely charged particles never stop moving apart as there will alwys be some repulsion
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Expert Comment

ID: 33430198
>> oppositely charged particles never stop moving apart as there will always be some repulsion

Yes, forever moving apart unless there are other forces to counter this. (I don't understand this dark matter and dark energy that is playing surprises in the universe.)

As two electrons move very very far apart, the repulsive acceleration becomes almost 0. When that happens, although still accelerating away from each other, they are essentially moving at almost constant speed. There is a negligible gravitational force that is overshadowed by the electrical force.
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Author Comment

ID: 33430201
Sorry, I don't get it.

If C is to the left of A then A is pushed to the right from the repulsion from C. A is also pulled to the right from the attraction from B. How can there be an equilibrium when A is moving? A wouldn't stop moving until it was on the right of B somewhere.

C is repulsed by A and attracted by B. They have written it the wrong way round so i would imagine the rest of the answer is wrong
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Expert Comment

ID: 33430334
Let's talk about this simpler problem from what you describe. If A and B are anchored along the axis, but C is free to move, would it then make sense that C is to the left of A?

Good catch..
>> C must be located to the left of A so that the repulsion interaction with B is balanced by the attractive interaction with A.

Change that to:
C must be located to the left of A so that the attractive interaction with B is balanced by the repulsion interaction with A.

But they are still setting the forces equal to each other.

B's force on C is to the right.
A's force on C is to the left.

Set C to the left of A so that the two forces balance out.

=======

If you put C to the right of B, then not only is B closer to C but also B's charge is stronger than A; so therefore, B will have a greater attractive force on C than A will have a repelling force.
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Expert Comment

ID: 33430467
I'll be back tonight.
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Author Comment

ID: 33430585
ok. i will be waiting, stuck, unable to proceed until you shine your guiding light :)
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Expert Comment

ID: 33430613
where are you stuck?
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Expert Comment

ID: 33430776
Assume A and B are anchored. Put down C at x location (first to the left of A), and if you want, to the right of B. Set the forces on C to be equal and opposite. Solve for x. See if you get the right answer. Post the work and we'll review.
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Author Comment

ID: 33432459
How can you assume A and B are anchored? That was why I couldn't do the problem in the first place and why i don't see how you can put C on the left.

But i will do as you suggested and set teh forces the same and see what happens
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Accepted Solution

shaleesh earned 500 total points
ID: 33432463
Actually i could come up with four solution for x[one of it matches with the solution provided in web]. You can find as below:
++++
Charge  :

Qa=18nC
Qb=-27nC
Qc=?

Distance:

Dab=60 * 10^-2 m
Dac=?
Dbc=?

Fnet=Fbc+Fca

F=k*Q1*Q2/d2

Fbc=-Fca

Qb/Dbc2=-Qa/Dca2

If c is located to right of 'b' then Dbc=x and Dca=(.6+x)m

-27 * 10^-9/x2 = -18 * 10^-9/(.6+x)2
3/2 =x2/(.6+x)2
3(.6+x)2=2x2
3(.36 + 1.2x + x2)=2x2
x2 + 3.6x + 1.08 = 0

x= (-b+sqrt(b^2 -4ac))/2a (or) (-b-sqrt(b^2 -4ac))/2a

a=1
b=3.6
c=1.08

-3.6+sqrt([3.6]2 -4*[1]*[1.08])/2(1)=-3.6+2.07846/2=-.7608
-3.6-sqrt([3.6]2 -4*[1]*[1.08])/2(1)=-3.6-2.07846/2=-2.8392

sqrt(12.96 -8.64)=sqrt(4.32)=2.07846

x = -0.7608 m
x = -2.83923 m

If c is located to left of 'a' then Dbc=(.6+x) m and Dca=x m

-27 * 10^-9/(.6+x)2 = -18 * 10^-9/x2
3/2=(.6+x)2/x2
2(.36+1.2x+x2)=3x2
-x2+2.4x+.72=0

x= (-b+sqrt(b^2 -4ac))/2a (or) (-b-sqrt(b^2 -4ac))/2a

a=-1
b=2.4
c=.72

-2.4+sqrt([2.4]2-4[-1][.72])/-2=-2.4+2.93938/-2=0.53938/-2=-.26969 m
-2.4-sqrt([2.4]2-4[-1][.72])/-2=-2.4-2.93938/-2=5.33938/2=2.669 m

sqrt(5.76 + 2.88)=sqrt(8.64)=2.93938

x = -0.26969 m
x = 2.669 m
+++
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Author Comment

ID: 33432586
well i still couldn't solve it :)

I got to Qa/x^2 = Qb/(x+6)^2
= 18/x^2 = -27/(x+6)^2

and I couldn't solve that. I plugged it into a solver and culdnt get an answer that way either
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Author Comment

ID: 33432612
I forgot its nC - will look again
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Author Comment

ID: 33432643
i still couldn't solve it

shaleesh: - why are you using -18*10^-9 and not +18
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Author Comment

ID: 33432703
Ok, i see why you set them to the same sign - because the forces are equal in magnitude but opposite direction.

But the reason why i couldn't go this question in the first place is because i don't understand how you can assume A and B are anchored. I'm fine with the maths of the problem
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Expert Comment

ID: 33432707
Fnet= Fbc+Fca
Fbc=-Fca
IF in equillibrium consider Fnet = 0

So Fbc = -Fca
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Author Comment

ID: 33434094
phoffric: are you around tomorrow?
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Author Comment

ID: 33434161
another thought:

if 2 oppositely charged objects accelerate towards each other don't they collide in some way when they meet?
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Expert Comment

ID: 33434205
> if 2 oppositely charged objects accelerate towards each other don't they collide in some way when they meet?
They might.
If they have angular momentum, they could miss each other in an elliptical or hyperbolic orbit.
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Expert Comment

ID: 33434211
If the objects are extremely tiny, then quantum effects become dominant as the objects approach each other. If an electron and positron collide into each other, they won't just bounce off each other like two balls:
http://en.wikipedia.org/wiki/Electron%E2%80%93positron_annihilation
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Author Comment

ID: 33436064
phoffric, still there?

one last thing: why can we assume A and B are anchored? Is this actually possible or just a simplification for the sake of the question? I was also wondering if you were around this weekend as I will have a MILLION questions about electricity as I'm going to have another crack at understanding it. I nearly got it a month or so ago and then had to stop and do something else.

I'd happily pay you if we could set up some sort of tutoring arrangement as I seem to understand the way you describe things and it would probably be more effective for me in the long run than going round in circles on my own!
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Expert Comment

ID: 33438556
andije,
If I were to charge you a mere \$1/question, and as you say, you have a MILLION questions about electricity, well, then that would cost you one million \$ and would put me a too high tax bracket! And do you really have that kind of \$\$ (or Â£Â£)?

Although I used to tutor physics, I have forgotten it all, and out of interest on specialized topics, I am trying to relearn some things in Physics I. See, for example,
http://www.experts-exchange.com/Other/Math_Science/Q_26392999.html

Also, I think I would have trouble competing in tutoring online due to global online tutoring services as you see from these new releases:
http://www.tutorvista.com/press/overview.php#nbcnews
http://www.physics247.com/

In order for me to help you in Physics II, in all honesty, I would have to review that myself (although I was #1 in my class of 100), it amazes me how much I forgot! All that work in computers pushed the memories into a tiny corner somewhere (maybe my brain, but I think it may have dropped to another part of my body).

Re; anchoring - I thought that might be a good place to start. To explore other possibilities leads to some interesting (and long) discussions. Something to think about - are we talking about stable or unstable equilibrium. That is, if the placement were a billionth of a mm away from the theoretical point of stability, what would happen?

Re: placing C to the right of B. Remember, there are really two unknowns in this equation - the value of the positive charge and the placement. For example, if we started intuitively saying that we wanted B to remain stable, then if you made C = A and placed C to the right of B at a distance |BC| = |AB|, then certainly B would experience a zero net force. But, also intuitively, you can tell that A and C will not be in equilibrium.
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Expert Comment

ID: 33439070
So, you can setup an x-axis, and call A (or B) the origin. Then put C down at any value x=c, and assume that the charge on C is +K. Now we need to solve for K and c such that the net force on A, B, and C is 0.

This is a fair enough exercise problem. The answer may prove to be that there is no solution in which case you would have proven that equilibrium for all three charged particles is not possible.

Proving that something is true or false is a valid thing to do. If someone were to disprove General Relativity, that may be worthy of a thesis (or at least a footnote in some future journal). A H.S. physics teacher once said that a circular mirror allowed collected light to focus on a single point. I went home to prove this assertion. I ended up disproving it, and showed the teacher my proof. He said I was wrong. But I then got the head of the Math department to verify my proof. I did not get a thesis or a footnote out of this. But at least the teacher did not penalize me for contradicting him.
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Author Comment

ID: 33440206
I notice there aren't many people contributing to your question. :) Maybe its a little hard!

I've tried to find both a physics and a maths tutor now with no luck. People aren't interested because I'm in that no man's land of learning for my own self improvement and not following a nice little syllabus that they have taught before.

I've had to resort to videos on you-tube. Soon there will be no human contact. I'll probably marry a virtual human projection and have digital babies. If I'm lucky I won't even have to leave the house and will play virtual tennis like sharon stone in total recall. Perhaps we can play online then???
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Expert Comment

ID: 33442068
I really got a kick out of this too:
http://www.imdb.com/title/tt0258153/
http://g4tv.com/videos/5695/Interview-With-Star-of-S1m0ne/

>> I notice there aren't many people contributing to your question.
Too many chefs in the kitchen ruin the stew.

>> nice little syllabus that they have taught before
Probably because they think that one point builds upon another. It's called "shop talk", where one word or phrase connotes weeks of thought. It's a question of being on the same page when a word or expression is used. Without following a syllabus, they don't really know whether you really understand what is being talked about.

Being in limbo is very tough. It's one of the reasons that I self-study rather than take courses. In a course I would start daydreaming because I already knew a bunch, and then miss the crucial 5 minutes of something new.

The you-tube videos (e.g., academic earth, mit, india, etc.) provide a very sound classroom environment. If you go from start to finish and actually do the assignments from the university, then you can learn as much.

This forum would then be useful. You could have a problem with an assignment, show the work, and then get guidance.

I have tried to find answers on physics forums, and they only are willing to go so far. They do not give out "trade secrets" unless the student "magician" has almost mastered the subject.
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Author Comment

ID: 33449024
It's the level that is the problem for me. My level is far too basic for the online university courses. I found a chemistry 101 course that i think people with very little chemistry might do if they were a physics undergraduate perhaps but I can't find a something like a basic online physics course for say biology undergraduates with no physics.

I hunted out this post from a few months ago and your answers on there are excellent. I pretty much understand why voltage is constant across a wire when i read that.
http://www.experts-exchange.com/Other/Math_Science/Q_26288244.html
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Expert Comment

ID: 33451884
>> My level is far too basic for the online university courses.
You are a potential PhD candidate, so maybe you are underplaying your skill. But nevermind that. Recall that I tutored a commercial artist who wanted to become a doctor (because of her dominant lawyer husband). She was having trouble with college chemistry. But I dropped that after one hour and started teaching her 3rd grade arithmetic (something about what the period means in a decimal number). Two years later she was top of her class in college Calculus; and a year later she enter medical school. So, if you have similar goals, start where you are competent, and build up solidly from there.

Your questions all have the curiosity of someone who takes a deeper scientific viewpoint than most questions. So, if you want to be a solid scientist then you may wish to start at the beginning and end at the never-end. Alternatively, you could teach kids, but now you will be competing with iPhones and iPad games in the classroom.
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Expert Comment

ID: 33451890
If you need H.S. physics (online), then ask a question for the best resources, or take an adult evening course on the subject (we have that kind of stuff here). I thought I saw H.S tutorials in physics. I'm not sure I understand your goals. Is physics necessary for your PhD program, or is this out of a personal interest to learn?
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Author Comment

ID: 33453614
The physics is personal interest.  I tried an adult evening class in both maths and physics but i live in a small city and both were cancelled due to lack of numbers.

Teaching might actually be the best way for me to fulfill my own personal curiosity. If i do a phd i will study some tiny bit of biochemistry and be an expert in that but my interest now I am older seems to just be in just knowing a bit more about the way things around me work. So i can either be an expert in modelling the immune response pathways in cows or learn general chemistry and physics and electronics.

I am completely unbalanced in my knowledge. Loads of biology. Some chemistry. Next to no physics and probably after some hard work a reasonable level of maths.

I have just done a really basic physics course and passed fine. Even the elctricity bit. But I only passed the electricity bit i know what formulas to use and how to plug in the numbers. But for some reason I want to know what's really going on and not just be a number cruncher.

I''ve found an online physics course (AP Physics) so will work through that and see how I feel after.

In the meantime would you ever so kind and look at this post for me:

http://www.experts-exchange.com/Other/Math_Science/Q_26406602.html#a33453554

I really thought you got a build up of electrons at one end of the wire.
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Expert Comment

ID: 33453817
I'm pretty busy today, but I'll try to help if I have a minute. Many good Experts already in your other Q. You need to address the confusion, as clearly as you can.

>> So there isn't a build up of electrons and negative charge in the wire attached to the negative terminal of a battery?

Maybe when someone says this doesn't apply, it is because they are thinking of a build-up at both ends with no charge difference in between the ends; and as you know, that is not the case.
Maybe you can phrase your question there as "Since there is an electric field, then doesn't that electric field arise from a net charge difference". (Actually, I think "charge differential" might be better terminology - I'm a bit rusty on this topic, as I've mentioned.)
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Author Comment

ID: 33467982
I do appreciate the help of other experts but I  like how you describe things. It makes sense. Anyway, I'm taking your advice and doing an online course from start to finish. Maybe some bits from other topics will help em think in a more physics mindset which will help here
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Expert Comment

ID: 33468497
I was trying to help you learn how to learn. Of course, it doesn't hurt to get the right book that has the right answers in it. :)

It is important that you ask the right follow-up questions when you do not understand something.
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Author Comment

ID: 33489519
Is it useful to think of electrons in the wire moving because of the electric field generated by the potential difference between the 2 plates of a battery? I've been thinking in terms of the force on an electron in a wire based on its position in the electric field between the terminals of a battery and I'm not sure that way of thinking is helping me understand current electricity
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Expert Comment

ID: 33489584
>> If A and B are oppositely charged objects and placed along an axis, why don't they move towards each other? Perhaps they do.
They certainly do if initially they are both at rest. I believe that with high enough initial velocities, the two charged objects may never move towards each other. I believe there is a critical "escape" velocity where they will escape from each other forever, in which case their acceleration will diminish towards 0, and their speeds will asymptotically approach a constant speed. (I've simplified the scenario to exclude everything in the universe except for these two charged objects.)

>> If they did move towards each other, wouldn't the force between them then get bigger and bigger the closer they get? And then presumably they'd move together even faster.
Yes, but as d-glitch indicated, not too close. Then you have to use the laws of quantum physics to determine what happens.
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Author Closing Comment

ID: 33526318
thanks
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