Challenge Physics Problem: Electric Fields
3 point charges where q1=q3=5.00uC q2=-2.00uC and a=0.100m. Find the resultant force exerted on q3. u=micro
First to get steps and answers correct gets fan & medal.

- anonymous

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- anonymous

|dw:1439605074957:dw|
q1 and q3 is positive while q2 is negative.
Knows the steps and answers.

- anonymous

@Robert136 come here.

- anonymous

One of the easier electric field problem btw.

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## More answers

- anonymous

First chapter in electricity and magnetism course

- anonymous

Great I'll get that rolling once I write a paragraph in a sec

- anonymous

I gotta do write up for the inductance coil wire with use of ferromagnetism to control the traffic

- anonymous

Might as well the fok do it here so we can share and argue about it

- anonymous

Sure you do

- anonymous

|dw:1439605460241:dw|
on the hypotenuse.

- anonymous

Forgot to say it is a right triangle.

- anonymous

|dw:1439605827422:dw|
Forgot to add this.

- anonymous

The 2 legs are congruent.

- anonymous

Inductance have looping coil to detect the presence of metal materials which are mostly the constituents of vehicles. Change in the electromagnetic frequency above the coil installed beneath the ground makes this system quite an effective means to prevent traffic congestion from happening. Usually they are located underground up to a few meters from the stop line with respect to the green light so that turning off and on of the blue light and red light can be determined dependent on the traffic on adjacent roads. This technology has increased its use past a few years mainly due to the increased efficiency in increasing the flow of traffic as well as reducing unnecessary emission of gas to prevent gas from being released. This technology is a practical application of electromagnetism and will likely be used for the generations to come mainly due to the flexibility and cost effectiveness of operation associated with this technology.

- anonymous

That's a rough write up for application of inductance on the traffic that work in collaboration with adjacent roads and intersecting traffic lights so that the congestion as well as frustration with unnecessary wait can be avoided.

- anonymous

Pretty well written or do you suggest revision before submission to a course I am taking online?

- anonymous

Prob misworded that first sentence I should have said inductance coil looping system

- anonymous

Wow, that is a lot of unnecessary info.
Was going to say lots of bull$#*@

- anonymous

What do you think?

- anonymous

What the fok bro

- anonymous

Looks good for your resume.

- anonymous

I mean seriously be honest your turn to help me is strike me with honesty bro i did this in 2 minutes

- anonymous

trying to pull out items from my memory which is not so reliable

- anonymous

but i think I have good trivial knowledge to deal with crap like this

- anonymous

I am honest, bro

- anonymous

so my writing good enough packed with information that you don't care

- anonymous

Thanks for letting me know the physics behind this physics problem.
You are great at teaching me physics!

- anonymous

Learned a lot from your teaching.
Too bad it went nowhere.

- anonymous

What do you mean bro

- anonymous

this was my assignment

- anonymous

and I did it so it's my turn to beat your wit in the problem you proposed

- anonymous

No it is mine. My question is on this board.

- anonymous

Go back to your nonsense board

- anonymous

so we doing this challenge right

- anonymous

Right on.

- anonymous

F=qvBsin) Lorentz force equation is at work

- anonymous

Wrong already.

- anonymous

Wait what are known values please elaborate

- anonymous

the a,q1,q2,q3

- anonymous

from your question charges are known right I get it but hey your diagram looks like crap

- anonymous

I thought it was a magnet bar

- anonymous

Not telling you the concepts and unknown variables.

- anonymous

No just 3 point charges. q1 and q3 are positive
q2 is negative.

- anonymous

Hint: nothing to do with your inductance loop coil

- anonymous

hehe

- anonymous

Alright I am totally onto this one

- anonymous

I thought you were head of Physics AMA

- anonymous

No I just had a verbal stifle with my brother.

- anonymous

He got pissed at me for being told not to sit on his retricewatching TV and he nearly through his glass at me so I had to give him a lecture

- anonymous

This is easiest electricity challenge problem I will post.
Now can you help me with more in the future?

- anonymous

threw*

- anonymous

Definitely my pleasure to help you in all respect

- anonymous

but a=0.1mm you are referring to acceleration correct?

- anonymous

Or did I prove myself a dork again

- anonymous

Why not give me a free lecture on my topic? ( I know how to solve it, but not much on electric fields)
I asked for one.
He threw glass at you and you gave him free lecture on what he needs?

- anonymous

No, it is m which is distance.

- anonymous

Alright I can give yo a free lecture no problem

- anonymous

I told him to fok off and if he does it again that I'd kill him

- anonymous

he's such a punk

- anonymous

on electric field only though.
No inductance,current,resistance.

- anonymous

Alright as you know electric field is the field surrounding a charged particle right

- anonymous

no ohm, voltage, etc.
Just motions in an electric field, electric field lines, electric field in a continuous charge distribution.

- anonymous

Yes.

- anonymous

Electric field only concerns with the amount of electrons

- anonymous

on one significant other because protons are assumed

- anonymous

and usually protons are not movable unless extreme force is used

- anonymous

you know until the point?

- anonymous

Unless extreme electric force?

- anonymous

protons can't generally be removed so just saying the problem is dominated by the amount of electrons in a given system

- anonymous

If they cannot be moved, then how come electric field is not dominated by protons?

- anonymous

flow of electrons or charged objects(increase in electrons is another way of saying it since protons are immobile) creates a type of electromagnetic force

- anonymous

charged objects implies that electrons are packed more than the object original possessed you get that

- anonymous

Is current speed the rate of electrons speed?

- anonymous

That also concerns with electrical potential difference you remember?

- anonymous

Nope. Lets go back to this question.
Problem first!
Concepts last!

- anonymous

Do you prefer using Skype as medium of communication?

- anonymous

Typing goes so slow

- anonymous

But really.
I'll give you a hint
There only 2 concepts to this problems and they are very basic.
Repulsive and attractive force and Coulomb Law.

- anonymous

Oh yea that's pretty elementary

- anonymous

Elementary concepts are sometimes the most tricky and confusing.
Now try it.

- anonymous

##### 1 Attachment

- anonymous

just for our reference

- anonymous

I did say this is one of the easier electric field problem.
This is 1st chapter out of 12 chapter in the E&M course.

- anonymous

Ok this is a cheese

- anonymous

Now solve for the resultant.

- anonymous

wait a moment I am looking up the k value because my memory is crap

- anonymous

Getting the resultant is a bit complicated.

- anonymous

what's the k again?

- anonymous

9.0 x 109 N â€¢ m2 / C2 Can I assume this since it's' in air

- anonymous

yes.

- anonymous

How would you calculate the resultant force then.
You can just explain the steps.

- anonymous

Alright no problem

- anonymous

I would first plug in the values in the equation for the electric field between q1 and q3 and then calculate another between q2 and q3 after which I will use the trigonometry to find the vector of each in regards to q2

- anonymous

You mean find the electric force between q1 and q3? Then find electric force between q2 and q3

- anonymous

Not quite right on the trig part.

- anonymous

Yeah exactly based on the q3 I would find the electric field acting from two different directions and use the cosine law to account for the angle because they are not in straight line.

- anonymous

With everything else aside this seems pretty simple when actually doing and less pondering.

- anonymous

Enthrall me now

- anonymous

You said I will use trig to find vector of each in regards to q2.
q2 is negative and q3 is positive.
The q3 is moving away from the q2 toward the right side in a straight line.
Therefore you use trig from the electric force between q1 and q3

- anonymous

q1 and q3 repel diagonal.
Use trig to figure the angle value and use it solve for x and y components of the q1-q3 electric force.

- anonymous

Show me the equation then?

- anonymous

Oh you are saying that we make a triangle and then bisect the base of the line connecting q1 and q3 so that properties will the identical to each other between q1 and q3 altogether attracting the -2qC now I get it that 's much better

- anonymous

the find the length of the line bisecting the triangle and that's basically the force acting towards q2 which then I do the calculation for the two and get the final answer!!!!

- anonymous

Not sure what you mean.

- anonymous

See thesis why I love physics the more you think about it the better your understanding becomes

- anonymous

ok I'll draw it for you

- anonymous

|dw:1439609545525:dw|

- anonymous

acting on the q2

- anonymous

because both q1 and q3 have all equal properties this way now that the line is bisecting the triangle.

- anonymous

Use the cosine law to find the bisecting line and that's the net electric force towards q2

- anonymous

and you would do another round with q2 and net q1 q3 to get the final answer.

- anonymous

|dw:1439609508807:dw|

- anonymous

C(cos45) to find x component
C(sin45) to find y component

- anonymous

There is one more step to this?
Which is it?

- anonymous

It is a right triangle with congruent legs, so it is at 45 degrees.

- anonymous

exactly like I drew

- anonymous

q2 to q3 is a straight line.

- anonymous

One way or the other you would need to extend the lines to account for the force acting against it using your trigs

- anonymous

My way is one way I think

- anonymous

bisect the triangle like you said because of the congruent legs, and given the same properties of q1 and q3 you can use the cosine alone to find the bisecting line equidiving the angle at q2

- anonymous

|dw:1439610086603:dw|

- anonymous

q2 and q3 attract in a straight line.
q1 and q3 repel diagonally.

- anonymous

You can do this only with the electric charges with same properties

- anonymous

Your graph is messy.
Use the straight line and text feature.

- anonymous

against one attracting

- anonymous

Do it and tell me the answer then.

- anonymous

Ok

- anonymous

|dw:1439610270416:dw|

- anonymous

Do you get my reasoning?

- anonymous

So this way you would need to know cosine(45)=adjacent/hypotenuse and I believe that hypotenuse is known so that way I can get the length of adjacent. Now you can either convert that to the ratio between hypotenuse(electrical field originally calculated) and adjacent(bisector) to get the Fnet of the q1 and q2 acting on q3.
Then Fe=(k*Fnet(q1,q3)(q2))/square of the length of bisector=electric charge acting on q2!

- anonymous

How would you know the charge of q4.
And it takes a while to figure out the length of q4 using pythagorean theorem

- anonymous

Why not just identify all the forces, then subtract if it goes left and add if it goes up and to the right.

- anonymous

q4 is valid because both electric fields have identical properties acting on the opposite charge at same angle achieved y the bisector

- anonymous

You can assume one of the congruent sides to be the electric field of either q3 or q2 and treat it as a side and get the bisector from that value. And that bisector which is smaller than either of them ( as far as I can see from the diagram ) the only thing will be the straigt electric field acting on the q2.

- anonymous

I think there is pythegorena ways to do it, but assuming one of the congruent sides as being equal to the calculated electric field of either q3 or q1 is good enough to get the corresponding bisector

- anonymous

This will eliminate having to use the ratio

- anonymous

Tell me the charge I would have to solve for find the resultant force?

- anonymous

|dw:1439611447511:dw|

- anonymous

and where would you find the electric force to use cos(45) at?

- anonymous

Ouch flip that upside down sorry |dw:1439611674098:dw|

- anonymous

Show me your forces and label them.

- anonymous

you follow my reasoning?

- anonymous

Write me the formula by labeling things and assuming.
Example: Let C be electric force between q1 and q2
Let B be electric force between q1 and q3

- anonymous

|dw:1439612004309:dw|

- anonymous

|dw:1439612083672:dw|

- anonymous

It's actually well founded

- anonymous

Relies on easier concepts

- anonymous

What point are you trying to find? I am confused.

- anonymous

q2 and q4?

- anonymous

no once q4(artificially made) is established no worries about anything else

- anonymous

for it will be q4 and q2 with opposite charges

- anonymous

Oh I forgot to write this. Find the resultant force on q3 -_-

- anonymous

q3?

- anonymous

I thought you said q2 but sure I'll work my way around it

- anonymous

You only want the resultant force on q3.
so q3 will rely on q1 and q2
Forget the force betweem q1 and q2

- anonymous

But are you convinced with my line of reasoning to find q2?

- anonymous

How are you finding q2?
With cos45?

- anonymous

I just said, the two charges with identical properties are acting on q2, hence attracting it. In this case exclusively the method I just told you about works in the resultant force acting on q2 if you read my drawing

- anonymous

the y component should be going up. Not down.

- anonymous

y component you mean q2?

- anonymous

*two opposite charges attract

- anonymous

Yes, it looks going down.

- anonymous

The answer has it that it goes up.

- anonymous

There is nothing pulling it downwards.

- anonymous

q2?

- anonymous

Opposite charges attract, same charges repel. They are different charges attracting

- anonymous

Are they no

- anonymous

If you back to the original problem.
What is the resultant force on q3?
|dw:1439612814025:dw|
All positive on the y axis.

- anonymous

You can try to debunk my explanation using your "own" theory and not just from your text book. Where is your support using princ

- anonymous

ciples

- anonymous

Alright my comp's dying so let me recharge it up.

- anonymous

Um, there is no other way to do bro.
You cannot assume things work geometrically by adding a test charge.
You cannot bisect something if the q1 and q2 are not even equal to each other.

- anonymous

Just calculate your answer.

- anonymous

It does not work at different charges.

- anonymous

==q1=q2 from your statement i said repeatedly "if two be the same

- anonymous

be back in a s

- anonymous

q1=q3
q2 is different.

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