At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
|dw:1439605074957:dw| q1 and q3 is positive while q2 is negative. Knows the steps and answers.
@Robert136 come here.
One of the easier electric field problem btw.
First chapter in electricity and magnetism course
Great I'll get that rolling once I write a paragraph in a sec
I gotta do write up for the inductance coil wire with use of ferromagnetism to control the traffic
Might as well the fok do it here so we can share and argue about it
Sure you do
|dw:1439605460241:dw| on the hypotenuse.
Forgot to say it is a right triangle.
|dw:1439605827422:dw| Forgot to add this.
The 2 legs are congruent.
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.
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.
Pretty well written or do you suggest revision before submission to a course I am taking online?
Prob misworded that first sentence I should have said inductance coil looping system
Wow, that is a lot of unnecessary info. Was going to say lots of bull$#*@
What do you think?
What the fok bro
Looks good for your resume.
I mean seriously be honest your turn to help me is strike me with honesty bro i did this in 2 minutes
trying to pull out items from my memory which is not so reliable
but i think I have good trivial knowledge to deal with crap like this
I am honest, bro
so my writing good enough packed with information that you don't care
Thanks for letting me know the physics behind this physics problem. You are great at teaching me physics!
Learned a lot from your teaching. Too bad it went nowhere.
What do you mean bro
this was my assignment
and I did it so it's my turn to beat your wit in the problem you proposed
No it is mine. My question is on this board.
Go back to your nonsense board
so we doing this challenge right
F=qvBsin) Lorentz force equation is at work
Wait what are known values please elaborate
from your question charges are known right I get it but hey your diagram looks like crap
I thought it was a magnet bar
Not telling you the concepts and unknown variables.
No just 3 point charges. q1 and q3 are positive q2 is negative.
Hint: nothing to do with your inductance loop coil
Alright I am totally onto this one
I thought you were head of Physics AMA
No I just had a verbal stifle with my brother.
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
This is easiest electricity challenge problem I will post. Now can you help me with more in the future?
Definitely my pleasure to help you in all respect
but a=0.1mm you are referring to acceleration correct?
Or did I prove myself a dork again
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?
No, it is m which is distance.
Alright I can give yo a free lecture no problem
I told him to fok off and if he does it again that I'd kill him
he's such a punk
on electric field only though. No inductance,current,resistance.
Alright as you know electric field is the field surrounding a charged particle right
no ohm, voltage, etc. Just motions in an electric field, electric field lines, electric field in a continuous charge distribution.
Electric field only concerns with the amount of electrons
on one significant other because protons are assumed
and usually protons are not movable unless extreme force is used
you know until the point?
Unless extreme electric force?
protons can't generally be removed so just saying the problem is dominated by the amount of electrons in a given system
If they cannot be moved, then how come electric field is not dominated by protons?
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
charged objects implies that electrons are packed more than the object original possessed you get that
Is current speed the rate of electrons speed?
That also concerns with electrical potential difference you remember?
Nope. Lets go back to this question. Problem first! Concepts last!
Do you prefer using Skype as medium of communication?
Typing goes so slow
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.
Oh yea that's pretty elementary
Elementary concepts are sometimes the most tricky and confusing. Now try it.
just for our reference
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.
Ok this is a cheese
Now solve for the resultant.
wait a moment I am looking up the k value because my memory is crap
Getting the resultant is a bit complicated.
what's the k again?
9.0 x 109 N • m2 / C2 Can I assume this since it's' in air
How would you calculate the resultant force then. You can just explain the steps.
Alright no problem
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
You mean find the electric force between q1 and q3? Then find electric force between q2 and q3
Not quite right on the trig part.
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.
With everything else aside this seems pretty simple when actually doing and less pondering.
Enthrall me now
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
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.
Show me the equation then?
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
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!!!!
Not sure what you mean.
See thesis why I love physics the more you think about it the better your understanding becomes
ok I'll draw it for you
acting on the q2
because both q1 and q3 have all equal properties this way now that the line is bisecting the triangle.
Use the cosine law to find the bisecting line and that's the net electric force towards q2
and you would do another round with q2 and net q1 q3 to get the final answer.
C(cos45) to find x component C(sin45) to find y component
There is one more step to this? Which is it?
It is a right triangle with congruent legs, so it is at 45 degrees.
exactly like I drew
q2 to q3 is a straight line.
One way or the other you would need to extend the lines to account for the force acting against it using your trigs
My way is one way I think
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
q2 and q3 attract in a straight line. q1 and q3 repel diagonally.
You can do this only with the electric charges with same properties
Your graph is messy. Use the straight line and text feature.
against one attracting
Do it and tell me the answer then.
Do you get my reasoning?
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!
How would you know the charge of q4. And it takes a while to figure out the length of q4 using pythagorean theorem
Why not just identify all the forces, then subtract if it goes left and add if it goes up and to the right.
q4 is valid because both electric fields have identical properties acting on the opposite charge at same angle achieved y the bisector
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.
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
This will eliminate having to use the ratio
Tell me the charge I would have to solve for find the resultant force?
and where would you find the electric force to use cos(45) at?
Ouch flip that upside down sorry |dw:1439611674098:dw|
Show me your forces and label them.
you follow my reasoning?
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
It's actually well founded
Relies on easier concepts
What point are you trying to find? I am confused.
q2 and q4?
no once q4(artificially made) is established no worries about anything else
for it will be q4 and q2 with opposite charges
Oh I forgot to write this. Find the resultant force on q3 -_-
I thought you said q2 but sure I'll work my way around it
You only want the resultant force on q3. so q3 will rely on q1 and q2 Forget the force betweem q1 and q2
But are you convinced with my line of reasoning to find q2?
How are you finding q2? With cos45?
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
the y component should be going up. Not down.
y component you mean q2?
*two opposite charges attract
Yes, it looks going down.
The answer has it that it goes up.
There is nothing pulling it downwards.
Opposite charges attract, same charges repel. They are different charges attracting
Are they no
If you back to the original problem. What is the resultant force on q3? |dw:1439612814025:dw| All positive on the y axis.
You can try to debunk my explanation using your "own" theory and not just from your text book. Where is your support using princ
Alright my comp's dying so let me recharge it up.
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.
Just calculate your answer.
It does not work at different charges.
==q1=q2 from your statement i said repeatedly "if two be the same
be back in a s
q1=q3 q2 is different.