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Electric Forces and Electric Fields

  1. Feb 17, 2004 #1
    Hello all,
    Physics is generally pretty hard for me, so understanding Electric forces has been a monster. I've done what I "know" how to do on the following problem, but I need some input from the professionals =) Thank you in advance.

    Two point charges, Q1 and Q2 are located a distance .20 meter apart. Charge Q1= +8.0 (Mu)C. The next electric field is zero at point P, located .40 meter from Q1 and .20 meter from Q2.

    1. Determine the magnitude and sign of charge Q2.

    I don't know how to get the answer, but I realize the magnitude is the force acting on Q2. So, F=k(q1 x q2)/(r^2). The total force acting on Q2 would have to include both P and Q2, so F=k(Q1Q2)/(.2^2)+k(Q2P)/(.2^2).
    Since the E field around P is 0..
    E=k x (P)/(r^2)
    0=k x q/(.2^2) + k x q/(.4)^2
    I'm clueless on getting the actual force, I only know Q1 and I don't know P. What can I do here?

    2. Determine the magnitude and direction of the net force on charge Q1. No idea.

    3. Calculate the electrostatic potential energy of the system.

    qV=UE V=k x (q/r)
    q1 x k x (q1/r) + q2 x k x (q2/r) + P x k x (P/r)

    The rest on this part is dependent on above.

    4. Determine the coordinate of the point R on the x-axis between the two charges at which the electric potential is zero.

    Again, I don't know what to do.

    5. How much work is needed to bring an electron from infinity to point R, which was determined in the previous part?

    Dependent on above.

    Even though this looks like the work of a slouch, it's really my best effort. I, again, thank you for any assistance you provide.
     
  2. jcsd
  3. Feb 18, 2004 #2
    I am learning this stuff right now too, so here is my shake at it...

    1. Start with a diagram. I put Q1 which is the 8(uC) particle on the origin, and put the unknown Q2 to the right, on the x-axis, .2m away. Now the problem states that there is a point with zero field (P), that is .4m from Q1, and only .2m from Q2. This puts P at .4m on the x-axis. So we have Q1 Q2 P at equal intervals.

    Next realize that for P to be zero E, then the fields from Q1 and Q2 must cancel each other. We know Q1 is +, so we can figure Q2 to be -.

    E1 + E2 = E
    E1=ke(Q1/r1^2) and E2=ke(Q2/r2^2)
    Where r1 and r2 are the distances of Q1 and Q2 from the know point P.
    The way to find a distance when E is zero...

    E1 + E2 = 0
    E1 = -E2
    ke(Q1/r1^2) = ke(Q2/r2^2)
    (Q1/r1^2) = (Q2/r2^2)
    Q2*r1^2 = Q1*r2^2

    We know all the variables except for Q2, so plug in and solve.
    I got -2.00e^-6 (C)

    2. Once you know the value of Q2, to find the force on Q1:
    F=ke(Q1*Q2)/r^2 should do it. Since the charges are opposite, the force will be in the direction of Q2. My answer is 3.596 N/C

    3. I believe the formula here is U=ke(Q1*Q2)/r. Note that r is not squared here. My answer is .7192 (N*m)

    Up to now I feel good about my answers for questions 1-3, although I am not 100% sure. I am shaky on questions 4 & 5.

    4. I think that V=ke(q1/r1)+ke(q2/r2) where r1 and r2 are the distances from Q1 and Q2 to some point in between them. I let r1=x and r2=(.2-x) since we know the distance between them is .2m.
    So...
    V=ke(q1/r1)+ke(q2/r2)
    0=ke(q1/r1)+ke(q2/r2)
    (q1/r1) = -(q2/r2)
    (q1/x) = -(q2/(.2-x)) subbed in my r's in terms of x
    Now solve for and sub in all the known values.
    My answer is x=.16m, that will be to the right of Q1, closer to Q2.

    5. I think we need to use W=ke(Q1/Q2)/r, which is the same as part question #3. However, this will be negative since it is work that was done in order to bring the new charge to the point.

    I am not so sure about this one, so I hope someone else will be kind enough to check all this work. I would really like to know if I am on track with all this.

    Let me know what you think too. Post your solutions or additional questions.
     
  4. Feb 18, 2004 #3
    Part 3 Anyone Jump in

    I've worked through part 1-2, works great. For part 3, U=ke(Q1Q1)/r.

    I've tried a number of variations on this formula. My original thought to find the answer would be:

    (8.99x10^9 x -2 x 8)/.2=-7.192 x 10^11 (N*m)

    So, the engima is twofold:
    1. Is the answer positive or negative?
    2. Would the answer be .7192 or -7.192 x 10^11?

    I'll be back in 1/2 hour to review part 4-5. Thanks for responding to my post.

    Anyone else, please feel free to jump in =)
     
  5. Feb 18, 2004 #4
    OK, for number 3, the answer should negative.
    This is because they are opposite charges, and so they attract. Since they attract, you would actually have to perform work to separate them, and so potential energy is negative. That is how I understand it.

    For the actual value of the answer, I think you made a calculation error.


    U=(8.99*10^9) * (8*10^-6) * (-2*10^-6) / .2 = -.7192

    Remember that Q1 and Q2 are micro coulombs, so they are to the 10e-6.

    Do you have an answer in your book for number 3?

    As for 4 and 5, we just covered some of that in class tonight, so maybe I can look at it fresh and come up with some more insight as to whether or not they are correct.
     
  6. Feb 18, 2004 #5
    Parts 4 and 5

    I'm still having difficulty with Part 4. With 5, however, the work should be positive I believe because it's being brought into the system, a negative work would mean it was attracted to the other particles. A positive work would mean it was repeled slightly, but work overcame it to bring it in at point R. I'm not sure though, if it's feasible to say W=ke(Q1Q2)/r because W=Fd so..
    Fd=ke(Q1Q2)/r distances r and d are the same I believe, so r=d
    F=ke(Q1Q2)/(r^2) This really doesn't leave us anywhere though, I don't think at least.

    On the other hand..

    (Change in V)=U/q=W/q

    So the q's cancel out and
    U=W
    Hence, I wonder if we can substitute U with W like you said. Just conjuring up ideas here.

    If we use your original method, why would the value .7192 N*m really be the work? I just don't see the justifiable "conceptual" reason why, but I could be missing it.

    Student's working together for a better world of Physics =)
     
  7. Feb 19, 2004 #6
    Alright, so I think we agree now on questions 1, 2 and 3.

    For #4, is the answer correct? Can you verify the answer in your book? I feel very good about it after reviewing it again.

    As for #5, I stand by my idea that the answer is simply the opposite of the potential energy. I am 99% sure I read that somewhere, but I cannot for the life of me find it again. If I do, I will quote the text to you.
    Also, remember I changed my answer for #3 from + to -. That means that 5 should be positive .7192J. My logic is that if the potential energy is -.7192J (it would require that much work to pull the charges apart so they no longer interact), it would take an equal magnitude, but opposite direction to push them together.

    Unless, I just thought that maybe we are being thrown off a little since the question asks about bringing a particle to point R. That is where we calculated the E potential to be zero. I think what I solved in question 5 is to bring the two charges Q1 and Q2 together from infinity.

    Along those lines, if the point R has zero electric potential, maybe the answer is zero work to bring in another particle to that point? Maybe I am just running off on some wild tangent?

    Let me know if these answers match up with your text book answers, if they are listed. That will give me a better idea of what to focus on. After spending all this time, I would really like to know the correct outcome.

    Anyone else can jump in here to help us out. Please...!!!???
     
  8. Feb 19, 2004 #7

    turin

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    Homework Helper

    Well, I'm going to come in and agree with the result of your tangent. This is basically a concept question, as opposed to a calculation question. From what I've read, point R is at zero electric potential. By convention (the issue of convention can trip you up sometimes), what is the electric potential an infinite distance away from the charges? Remebering that potential energy is arbitrary do to an arbitrary offset, the convention sets the offset. Then, assume that the charge is brought to the point, R, with just the right amount of energy to sit there motionless (no kinetic energy at R; an extremely subtle, but physically significant issue that can also cause confusion). "Calculate" (you will see that the calculation is quite trivial) the change in mechanical energy to calculate the work. The mechanical energy is entirely electrical potential energy. One last important point: this calculation works because the electrostatic field is conservative.

    Here's a more intuitive/qualitative understanding:
    There is a certain amount of work pushing the electron "up the hill" from infinity. But, when it passes the "top of the hill," it will give this work back as it "rolls down the other side" of its own accord. Or, coming from the other side, the electron rolls down into a valley, and, at the bottom it has enough kinetic energy to get it back up out of the valley on the other side.
     
    Last edited: Feb 19, 2004
  9. Feb 19, 2004 #8
    Turin,
    Your right about one thing...confusion.

    If I understand you, then the energy required will be zero due to some sort of cancellation?
    You mention the calculation of this being trivial. Can you help us out with that?

    Zeus - Any insight on the correct answers?
     
  10. Feb 19, 2004 #9
    Part 5

    I'm not actually sure on this part. I won't be meeting until tommorrow, so I'll ask then. I was talking to a friend of mine and he said he got 0 (he is quite the Physicist too), but he didn't have time to explain.

    Sorry for the lack of info, I'll give an update when I get more info.
     
  11. Feb 19, 2004 #10

    turin

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    I suppose you could say it that way.




    Yes. The formula invokes energy conservation:

    W = ΔEmech,

    where W is the work done on the system (the charge being brought in) and Emech is the mechanical energy of the system (kinetic energies and potential energies). To write it out a bit more explicitly:

    W = Emech,final - Emech,initial

    = (KEfinal + PEfinal) - (KEinitial + PEinitial)

    = (0 + 0) - (0 + 0)

    = 0,

    where the kinetic energies are set to zero BY ASSUMPTION, the final potential energy is given to be zero in the problem, and the initial potential energy is set to zero BY CONVENTION (a convention which I believe you already used in parts 3 and 4).

    Another thought that I just had that might help alleviate the ambiguity is to justify the assumption that KEfinal = 0. If KEfinal > 0, then more work would be required (this can be seen by inspecting the work energy formula above). And, in classical physics (the stuff you're doing), kinetic energy cannot be negative. So, you can sort of interpret the question as, "What is the minimum amount of work out of all possible motions to get the charge to point R from infinity?" This takes care of setting KEfinal = 0. Unfortunately, the KEinitial = 0 will have to remain ambiguous. In cases such as these, the rule of thumb for assigning uspecified parameters is to assign them in such a way that makes the problem as simple as possible without making it trivial. Of course, it is trivial in this case, but that's generally not unusual for a concept question (there's always an exception to the rule, even in physics).
     
    Last edited: Feb 19, 2004
  12. Feb 19, 2004 #11
    Thanks Turin for spelling out your response to #5. I have a better feel for it now, but still not a complete understanding. Maybe some more thought and reading will help it sink it in.

    I am curious to know what you think about parts 1-4. Should I assume that it all looks correct so far since you didn't mention anything negative about it? I guess we will find out for sure once Zeus' class meets tomorrow. Unless Zeus has one of those professor's who let's things lag on for weeks without solutions.
     
  13. Feb 19, 2004 #12
    Please Look at Other Post

    We'll see Paul =) My professor is unpredictable, lol.
    Hey, Turin, thanks so much for your input into this problem! If you have a moment, could you look at my other post?

    I love these forums!
     
  14. Feb 19, 2004 #13

    turin

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    It looks correct to me. I didn't check numbers, but the plans of attack look sound.
     
  15. Feb 20, 2004 #14
    So Zeus, did today's class shed any light on this problem?
     
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