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Supposedly easy electromag question which I cannot do sadly.

  1. Jul 1, 2009 #1
    1. The problem statement, all variables and given/known data

    Point charges of magntitude q1, 2q1 and q2 are located at points (1,1,0), (2,0,0) and (-1,0,0). Find the relation between q1 and q2 so that the y-component of the total force on a unit positive charge at the point (0,1,0) is zero.

    2. Relevant equations

    Coulombs law.

    3. The attempt at a solution

    I know I have to resolve the vectors so that the final solution is zero. However I cannot do this. Please can you give me a method of how I might do this. Please note this is revision and not homework so you arn't helping me cheat. thanks
  2. jcsd
  3. Jul 1, 2009 #2
    Draw out all three of your points nice and big on a piece of paper. Use some trigonometry to find the y component of each of your forces on the fourth charge. You set this total charge, to zero, then find the relation you were asked for via some algebra. IE, you're looking for q1/q2
  4. Jul 1, 2009 #3
    If r1 is where a point charge is located, then the field from that charge at another position, r2, is given by E = kq/r^2 in the r-direction where r = r2-r1 is the separation vector that points from r1 to r2.

    So for each of your point charges you need to find r. Remember that r2 (the point where you want to know the field) will remain (0,1,0) throughout. For the first one it would be:
    r = r2-r1 = (0,1,0) - (1,1,0) = (-1,0,0)
    and so r^2 = r.r = 1
    This means that, for this first charge:
    E = k*q1/1 = k*q1 in the r direction
    And since r is already normalized, we can just multiply E by r to get the vector value for E (if it were not, we would need to normalize it so that we do not change the magnitude of E when we multiply by it - we only want to add the directional dependence!):
    E = k*q1(-1,0,0)

    Once you do this for each of your 3 charges, you can add of the three vectors you found for the fields at that point due to each of them to find the total field. Then set the y-component equal to 0 to find an equation that will give you your relation between q1 and q2.
  5. Jul 2, 2009 #4
    Thanks very much. I can thankfully now do that problem, however my hope that I could do similar problems seems unfounded.

    I have worked out that the direction is 45% in the postive x and y direction. However for the life of me I cannot get a correct answer for the other part. Do I have to use trigonometry??? I am at the moment trying to work out the three components of the E field and then sum them, however this is not really working.

    Any help really appreciated.
  6. Jul 2, 2009 #5
    Summing the contributions would surely work, show us what you did and we'll see what went wrong. (surely it will involve trigonometry *somewhere*)
  7. Jul 2, 2009 #6
    For any problem like this, you resolve it into components by remembering that in Coulomb's Law:
    All of the directional dependence is carried in
    Where r1 is the point where the charge is and r2 point where you want to find the field. This means that for each point you just need to calculate magnitude of E and then the unit vector r-hat. When you multiply them together, you get the full vector-value for E at that point from that specific charge. Then you just repeat with each charge and add the vectors together.

    For example, the +q charge in your problem would have:
    So now you can compute
    Now finding E is easy

    No need to compute percentages in any direction or anything. These simple vector quantities are a most acceptable answer format!
    Last edited: Jul 2, 2009
  8. Jul 2, 2009 #7
    Firstly thankyou for taking the time to explain all of that, it has been extremely useful. However I have a question. Once you have worked out the 3 components, do you simple sum them? If this is the case, how do you treat the unit vectors ie (1,1) part? Thanks
  9. Jul 2, 2009 #8
    Glad to help!

    That's a good question. Your absolutely right on the first part; once you have the contributions to the field at the origin from each of the three charges you just add them together to find the total field. Notice that I call these 'contributions' though; a component is something different. Each charge contributes to the field, and the field (and also each contribution) has multiple (2 in this case, since it's a 2-dimensional problem) components.

    A cool property of vectors is that, to add them, all you have to do is add their components. Let's look at a different problem (so that I don't rob you of the enjoyment of doing your problem). Say I have two point charges, on q and one 3q at (-1,0) and (0,-2) respectively and I'm interested in the electric field they produce at the origin.

    Now you can just read off the components:
    - In the x direction, the field is kq
    - In the y direction, the field is (1/2)kq

    And if you like percentages/ratios, you can see from the vector-part that the field is twice as strong in the x-direction as it is in the y-direction (66% x-direction, 33% y-direction), but the single vector quantity kq(1,1/2) says it all.

    Does that make sense?

    For the first one I have:
    so that
    and for the second we get:
    so that

    Finally, to find the total field at the origin, we add these two contributions from our two charge together by adding them component-wise, that is:
    [tex]\vec{E}_\textit{total}=\vec{E}_1+\vec{E}_2=kq[(1,0)+\frac{1}{4}(0,2)]=kq[(1,0) + (0,\frac{1}{2})] = kq(1,\frac{1}{2})[/tex]
  10. Jul 2, 2009 #9
    Yes that does make sense I think. However in the part which I have put in bold do you mean just another q or 3q?

    Also as my answer is still not coming out as what is in the solution how do you resolve \vec{E}_\textit{total}=\vec{E}_1+\vec{E}_2=kq[(1,0)+\frac{1}{4}(0,2)]=kq[(1,0) + (0,\frac{1}{2})] = kq(1,\frac{1}{2})? I multiplying through by this... However my answer comes out at around 9000V/m rather than the required 6000ish...
    Last edited: Jul 2, 2009
  11. Jul 3, 2009 #10


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    Homework Helper

    The charge q1 is on the x-y plane, the charge 2q1 is on the x-axis and the charge q2 is on the negative x-axis. The unit positive charge is on the y-axis.
    The force due to q1 on the unit positive charge is parallel to the x-axis and hence it has no y component.
    To have zero y-component 2q1 and q2 must have opposite signs.
    The distance between 2q1 and the unit positive charge is (5)^1/2 and distance between q2 and unit positive charge is (2)^1/2. Find the forces due to these two charges on unit positive charge. Find the components of these forces along the y-axis.
    Equate them to get the ratio of the charges.
  12. Jul 3, 2009 #11
    Sorry about that. I was just trying to make the problem different than yours so that you could see how I worked this one and then you would still be able to do yours, but it looks like I forgot about the 3q as soon as I started computing. So you're right, the problem I did was with both charges only being q. Still, they are in different positions than the ones in your problem, you you'll need to compute those using the same process (just with your positions/charges instead of the ones that I did).

    In your problem with charges of -q at (a,0) and (0,a), what do you get when you compute the contributions to the electric field at the origins from each charge? Remember:

    So for the first one you will have:
    [tex]\vec{r}=(0,0)-(a,0) = (-a,0)[/tex]

    You can work out the r-vector for the second charge and then you can stick each of these into E = kq/r^2 r-hat to find their contributions (remember to normalize r to get r-hat!). The just add them component-wise.

    (I probably made my example problem look too much like yours - my fault, I didn't mean to confuse you, but I can see how that would be pretty confusing).
  13. Jul 3, 2009 #12
    Thanks. You have been amazingly clear actually. I feel like I am starting to understand these questions now. Infact I can get the answers, I really appretiate all of the help you have given me and apologise if I was slow on the uptake at times.
    Last edited: Jul 3, 2009
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