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anonymous
 one year ago
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
 one year ago
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.

This Question is Closed

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1439605074957:dw q1 and q3 is positive while q2 is negative. Knows the steps and answers.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@Robert136 come here.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0One of the easier electric field problem btw.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0First chapter in electricity and magnetism course

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Great I'll get that rolling once I write a paragraph in a sec

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I gotta do write up for the inductance coil wire with use of ferromagnetism to control the traffic

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Might as well the fok do it here so we can share and argue about it

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1439605460241:dw on the hypotenuse.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Forgot to say it is a right triangle.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1439605827422:dw Forgot to add this.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0The 2 legs are congruent.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Inductance 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
 one year ago
Best ResponseYou've already chosen the best response.0That'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
 one year ago
Best ResponseYou've already chosen the best response.0Pretty well written or do you suggest revision before submission to a course I am taking online?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Prob misworded that first sentence I should have said inductance coil looping system

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Wow, that is a lot of unnecessary info. Was going to say lots of bull$#*@

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Looks good for your resume.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I mean seriously be honest your turn to help me is strike me with honesty bro i did this in 2 minutes

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0trying to pull out items from my memory which is not so reliable

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0but i think I have good trivial knowledge to deal with crap like this

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so my writing good enough packed with information that you don't care

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Thanks for letting me know the physics behind this physics problem. You are great at teaching me physics!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Learned a lot from your teaching. Too bad it went nowhere.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0What do you mean bro

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0this was my assignment

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0and I did it so it's my turn to beat your wit in the problem you proposed

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0No it is mine. My question is on this board.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Go back to your nonsense board

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so we doing this challenge right

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0F=qvBsin) Lorentz force equation is at work

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Wait what are known values please elaborate

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0from your question charges are known right I get it but hey your diagram looks like crap

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I thought it was a magnet bar

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Not telling you the concepts and unknown variables.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0No just 3 point charges. q1 and q3 are positive q2 is negative.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Hint: nothing to do with your inductance loop coil

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Alright I am totally onto this one

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I thought you were head of Physics AMA

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0No I just had a verbal stifle with my brother.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0He 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
 one year ago
Best ResponseYou've already chosen the best response.0This is easiest electricity challenge problem I will post. Now can you help me with more in the future?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Definitely my pleasure to help you in all respect

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0but a=0.1mm you are referring to acceleration correct?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Or did I prove myself a dork again

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Why 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
 one year ago
Best ResponseYou've already chosen the best response.0No, it is m which is distance.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Alright I can give yo a free lecture no problem

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I told him to fok off and if he does it again that I'd kill him

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0on electric field only though. No inductance,current,resistance.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Alright as you know electric field is the field surrounding a charged particle right

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0no ohm, voltage, etc. Just motions in an electric field, electric field lines, electric field in a continuous charge distribution.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Electric field only concerns with the amount of electrons

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0on one significant other because protons are assumed

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0and usually protons are not movable unless extreme force is used

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0you know until the point?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Unless extreme electric force?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0protons can't generally be removed so just saying the problem is dominated by the amount of electrons in a given system

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0If they cannot be moved, then how come electric field is not dominated by protons?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0flow 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
 one year ago
Best ResponseYou've already chosen the best response.0charged objects implies that electrons are packed more than the object original possessed you get that

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Is current speed the rate of electrons speed?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0That also concerns with electrical potential difference you remember?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Nope. Lets go back to this question. Problem first! Concepts last!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Do you prefer using Skype as medium of communication?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0But 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
 one year ago
Best ResponseYou've already chosen the best response.0Oh yea that's pretty elementary

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Elementary concepts are sometimes the most tricky and confusing. Now try it.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0just for our reference

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I 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
 one year ago
Best ResponseYou've already chosen the best response.0Now solve for the resultant.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0wait a moment I am looking up the k value because my memory is crap

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Getting the resultant is a bit complicated.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.09.0 x 109 N • m2 / C2 Can I assume this since it's' in air

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0How would you calculate the resultant force then. You can just explain the steps.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I 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
 one year ago
Best ResponseYou've already chosen the best response.0You mean find the electric force between q1 and q3? Then find electric force between q2 and q3

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Not quite right on the trig part.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yeah 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
 one year ago
Best ResponseYou've already chosen the best response.0With everything else aside this seems pretty simple when actually doing and less pondering.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0You 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
 one year ago
Best ResponseYou've already chosen the best response.0q1 and q3 repel diagonal. Use trig to figure the angle value and use it solve for x and y components of the q1q3 electric force.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Show me the equation then?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oh 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
 one year ago
Best ResponseYou've already chosen the best response.0the 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
 one year ago
Best ResponseYou've already chosen the best response.0Not sure what you mean.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0See thesis why I love physics the more you think about it the better your understanding becomes

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok I'll draw it for you

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1439609545525:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0because both q1 and q3 have all equal properties this way now that the line is bisecting the triangle.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Use the cosine law to find the bisecting line and that's the net electric force towards q2

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0and you would do another round with q2 and net q1 q3 to get the final answer.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1439609508807:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0C(cos45) to find x component C(sin45) to find y component

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0There is one more step to this? Which is it?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0It is a right triangle with congruent legs, so it is at 45 degrees.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0q2 to q3 is a straight line.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0One way or the other you would need to extend the lines to account for the force acting against it using your trigs

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0My way is one way I think

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0bisect 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
 one year ago
Best ResponseYou've already chosen the best response.0dw:1439610086603:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0q2 and q3 attract in a straight line. q1 and q3 repel diagonally.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0You can do this only with the electric charges with same properties

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Your graph is messy. Use the straight line and text feature.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0against one attracting

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Do it and tell me the answer then.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1439610270416:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Do you get my reasoning?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So 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
 one year ago
Best ResponseYou've already chosen the best response.0How would you know the charge of q4. And it takes a while to figure out the length of q4 using pythagorean theorem

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Why not just identify all the forces, then subtract if it goes left and add if it goes up and to the right.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0q4 is valid because both electric fields have identical properties acting on the opposite charge at same angle achieved y the bisector

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0You 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
 one year ago
Best ResponseYou've already chosen the best response.0I 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
 one year ago
Best ResponseYou've already chosen the best response.0This will eliminate having to use the ratio

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Tell me the charge I would have to solve for find the resultant force?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1439611447511:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0and where would you find the electric force to use cos(45) at?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ouch flip that upside down sorry dw:1439611674098:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Show me your forces and label them.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0you follow my reasoning?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Write 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
 one year ago
Best ResponseYou've already chosen the best response.0dw:1439612004309:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1439612083672:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0It's actually well founded

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Relies on easier concepts

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0What point are you trying to find? I am confused.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0no once q4(artificially made) is established no worries about anything else

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0for it will be q4 and q2 with opposite charges

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oh I forgot to write this. Find the resultant force on q3 _

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I thought you said q2 but sure I'll work my way around it

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0You only want the resultant force on q3. so q3 will rely on q1 and q2 Forget the force betweem q1 and q2

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0But are you convinced with my line of reasoning to find q2?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0How are you finding q2? With cos45?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I 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
 one year ago
Best ResponseYou've already chosen the best response.0the y component should be going up. Not down.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0y component you mean q2?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0*two opposite charges attract

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yes, it looks going down.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0The answer has it that it goes up.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0There is nothing pulling it downwards.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Opposite charges attract, same charges repel. They are different charges attracting

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0If you back to the original problem. What is the resultant force on q3? dw:1439612814025:dw All positive on the y axis.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0You can try to debunk my explanation using your "own" theory and not just from your text book. Where is your support using princ

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Alright my comp's dying so let me recharge it up.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Um, 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
 one year ago
Best ResponseYou've already chosen the best response.0Just calculate your answer.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0It does not work at different charges.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0==q1=q2 from your statement i said repeatedly "if two be the same

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0q1=q3 q2 is different.
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