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Electric Field

  1. Jul 7, 2006 #1
    Alright, so I seem to be having difficulty setting this one up. Here's the problem:


    The way I began to go about it is as follows...

    [tex]\Sigma \vec{E}=\vec{E_{1}}+\vec{E_{2}}+\vec{E_{3}}=0[/tex]




    It's the distance, I don't really know how to resolve that distance r... My reasoning is this... Because the 2nd charge on the x-axis, the one that's -7 micro-coulombs is larger than the 1st charge, 6 micro-coulombs, the added charge is going to be to the left of the 1st charge (6 micro-coulombs)

    So that's how I get what I got.

    Does this make sense? I'm not doing something right because after algebra (even computer algebra), I'm getting the wrong answer.

  2. jcsd
  3. Jul 7, 2006 #2
    The electric field at the origin due to the charges whose positions are known is constant. So figure out the magnitude and direction of that, then decide which side of the origin to put the third charge, and finally calculate the distance at which to put it. I'm not sure what you're doing with the x and the (x+6).
  4. Jul 7, 2006 #3
    Why are you using [itex]x[/itex] instead of numerical distances? Just plug in the numbers you are given.

    [tex]\vec{E_{1}}=\frac{kq_{1}}{r^2}=\frac{(9\times10^9) (6\times10^{-6})}{0.01^2}[/tex]
    [tex]\vec{E_{2}}=\frac{kq_{2}}{r^2}=\frac{(9\times10^9) (-7\times10^{-6})}{(0.05)^2}[/tex]
    [tex]\vec{E_{3}}=\frac{kq_{3}}{r^2}=\frac{(9\times10^9) (1\times10^{-6})}{(x)^2}[/tex]

    You also know that [tex]\vec{E_{at 0}}= \vec{E_{1}}+\vec{E_{2}}+\vec{E_{3}}[/tex]. Then use the principle of the superposition of fields.

    I just learned electric fields in class today, so I'm eager to test my knowledge :smile:.
  5. Jul 7, 2006 #4


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    You are confusing *vectors* and *magnitudes*. What you calculate on the right hand side are the magnitudes of the electric fields.

    (EDIT: It's even worse than this because I just noticed that you kept a minus sign in one of the charges so you were not even calculating the magnitudes!! What you should FIRST do is to calculate the MAGNITUDES of the E fields produced by each of the two charges and then proceed as I explain below)

    To find the *vectors*, you must determine the direction of the electric fields produced by the two given charges at the origin. *Now* you will be able to write the electric field vectors ( as multiples of the unit vector [itex] {\hat i}[/itex]) produced by the two charges at the origin and do a vector addition. If the two E fields are not in the same direction, adding the magnitudes of the two E fields to get the total E field would be incorrect.

    I am not saying that adding the magnitudes of the E fields of the first two charges is necessarily wrong in this problem, I am saying that you must be careful because if you don't make the distinction between vectors and magnitudes, you could be in big trouble in an exam.

    Then you can find what should be the magnitude of the unknown charge by setting the magnitude of its E field to the magnitude of th etotal E field produced by the first two charges and solving for x.
  6. Jul 7, 2006 #5
    Ahy. So I'm a bit confused. In another problem I had a bit of difficulty with something like this. I often get overwhelmed when I'm asked to deal with vectors.

    Because this is all along the x-axis, I only have to deal with x-components, and thus the entire vector is just the addition of the x-components.

    ...E1 is in the + direction, E2 is in the - direction, and E3 is in the + direction. As I indicated.

    I'm pretty sure I'm doing the vector stuff right, I just don't know how to solve for the distance.

    I'm pretty sure Saketh isn't correct, because from what I understand, when trying to find the electric field, I've usually dealt in absolute value distances. ...The two charges I was given the location of are 6 units apart, so that's the reason for the x+6...

    I don't know. I'm still awfully confused about setting this up correctly. It's the r's I'm having difficulty with!
    Last edited: Jul 7, 2006
  7. Jul 7, 2006 #6
    I always hated when I took physics thinking of only the x componenets, or only the y components, etc... Instead I just thought of everything as being three dimensional and setup all my problems like this. I also think using the [itex] \hat i,\,\,\, \hat j,\,\,\,\hat k [/itex] notation is ridiculous, and instead use the ahhh... forgot what it's called notation where a vector would look like:

    [tex] \vec r = \left[ \begin{array}{c} x_0 & y_0 & z_0 \end{array} \right] [/tex]

    So this might be better for you. I found that once I got used to writing it the "long" way, I could do the little short hand tricks (my opinion) that the book uses.

    I believe r is just the distance to the orgin (someone please correct me if I'm wrong).

    So, since you are only going to be dealing with three dimensional vectors. Lets say you have a point located at x = 5cm. Then the vector that represents position would be:

    [tex] \vec p_1 = \left[ \begin{array}{c} 5 & 0 & 0 \end{array} \right] [/tex].

    The [itex] r [/itex] is a scalar, and it represents a distance from the orgin. So this is done with the vector norm operation, ie:
    [tex] r = |\vec p_1| = \sqrt{(5)^2+(0)^2+(0)^2} = 5 [/tex]

    Also remember that:
    [tex] \vec E=\frac{kq}{r^2} [/tex]

    is actually a vector function, that you can write to take the x, y, and z components:

    [tex] \vec E(x,y,z)= \frac{kq}{\left( \sqrt{x^2+y^2+z^2}\right)^2} [/tex]
    Last edited: Jul 7, 2006
  8. Jul 7, 2006 #7
    r would be the distance from the origin??

    Is this correct?
  9. Jul 7, 2006 #8


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    Ok, so let's work thos one out step by step, ok?
    This is absolutely correct
    Wait. There is a problem here. How did you decide of the directions of the fields E1 and E2??
  10. Jul 7, 2006 #9
    ...Egh. Well, I figured that because q1 is positive, q2 is negative, and q3 is positive, that E would follow... But now that you mention that, I remember seeing a problem where the field was going opposite the direction of the charge. So would it be the other way around?

    Would E1 be -, E2 +, and E3 - ???
  11. Jul 7, 2006 #10


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    No, you are going about this the wrong way, I am afraid. You cannot get the direction of the E field at a point simply by looking at the sign of the charge! You also need to look at *where* the charge is located.

    The other thing is that you should not think about E3 for now! You will have to add [itex] {\vec E_1} + {\vec E_2} [/itex] before you can decide in what direction E3 must be pointing.

    Going back to E1 and E2. The rule you *must* follow is this:

    The E field produced by a positive point charge points away from the charge at all points.

    The E field produced by a negative point charge points toward the charge.

    Now, given the signs of q1 and of q2 and given their location, what is the direction of [itex] {\vec E_1} [/itex] and [itex] {\vec E_2} [/itex] at the origin?

    Then, what is the value of [itex]E_{1x}[/itex] and [itex] E_{2x} [/itex]?
    Then, what is the value of the sum of those two values?
    Then, what must be the direction and magnitude of E3?

    Once you have answered these, we will discuss the value and position of q_3.

  12. Jul 7, 2006 #11


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    Saketh was right to use 0.01 and 0.05 for the distances! If you want to find the E field produced by a point charge at certain point, the distance you use is the distance between the charge and the point. So that's 1 cm (so 0.01 m) for one charge and 0.05 m for th eother charge. No need for an unknown "x" for those two charges, you are given the distance between the charges and the point!!
  13. Jul 8, 2006 #12
    Okay... So this is what I have. Please let me know if I'm doing this correctly. I was under the impression that the electric field was in the same direction as the electrical force. Here's a diagram:


    And as for r, I'm taking that as a distance between the points. So r in both cases would be 6cm. (The answer is to be in cm, so I'm not converting it into meters)

    So I'd just plug that in, and whatever I get for that would provide the value for E3?... I'd take the sum of those two, and just make the negative of that sum E3?... Then solve for r?
  14. Jul 8, 2006 #13
    Right, but I don't have the distance between that point charge and my two points. I'm asked to find the location of a point with charge 1micro-C that makes E=0... What he's talking about is the location of the origin.
  15. Jul 8, 2006 #14


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    Unfortunately, I think you are confusing a lotf of things. sorry to tell you :frown:

    First, no it's not true that the E field is always in the same direction as the force on a charge.

    But that's irrelevant to this problem. Because in this problem, you do not need AT ALL to consider the forces on the charges q1 and q2!!!

    What you want is the E field AT THE ORIGIN! Where there is NO charge at all!

    What is the E field produced by q1 at the origin?
    (direction and magnitude)

    What is th E field produced by q2 at the origin?
  16. Jul 8, 2006 #15
    Looking back at this, it doesn't make sense to find E1 and E2 and then to factor in E3 after all is done with E1 & E2... The value of E1 & E2 depend on the distance from q1 & q2 to the location of the point I'm trying to find. So I would have an unknown in each of the denominators.
  17. Jul 8, 2006 #16


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    You do not need the distance between q3 and those two charges.
    I think you should reread carefully the question. We want the total E field at the origin to be zero! So we need to calculate the E field at the origin!!!!! And you do know the distance between the first two charges and the origin so you *can* get numbers for those two E fields at the origin!
  18. Jul 8, 2006 #17


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    Read the question carefully. You need to find the position of q3 such that the total E field is zero at the origin . So the distance between q1 and the point where you are calculating the E field, namely the origin is known!! And the distance between q2 and the point where you are calculating the E field, namely the origin, is also known! So you can calculate the numerical values of the E fields produced by q1 and q2 at the point! It's only when you will consider q3 that you will have an unknown because you don't know where q3 is.
  19. Jul 8, 2006 #18
    Okay. I'm really sorry about all of the misunderstandings. I got a lot of things crossed in my head, made me think I was looking for something other than what I was really looking for. Thank you, and sorry again.




    I'm calculating this in cm because the answer asks for cm. So... Just in case you were wondering...


    So I'm guessing that E3 would then be -51480N/C, and then solve for the missing r?
    Last edited: Jul 8, 2006
  20. Jul 8, 2006 #19


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    You made the same mistakes that you did in the other thread:redface:

    First, you should calculate the magnitude and directions separately. so, for the magnitudes, use the absolute values of the charges. And write E1 and E2 (no vector signs)

    Second, put the distances in meters.


    [tex]E_{1}=\frac{k |q_{1}|}{r^2}=\frac{(9\times10^9)(6\times10^{-6})}{.01^2}=5.4 \times 10^8N/C[/tex]

    [tex]E_{2}=\frac{k |q_{2}|}{r^2}=\frac{(9\times10^9)(7\times10^{-6})}{5^2}=2.52 \times 10^7 N/C[/tex]

    This gives you the *mangnitudes*. Now determine the directions. Once you know the directions and magnitudes, you may write the x components of each. Add them up (taking into account their sign).

    Then determine where a positive charge (since q3 is positive) would have to be located to cancel this (to the left or to the right of the origin?). Then find the distance it must be at so that the magnitude of E3 will be the same as the magnitude of [itex] {\vec E_1} + {\vec E_2} [/itex].

    I have to go soon (it's past 1 am here).

    Best luck
    Last edited: Jul 8, 2006
  21. Jul 8, 2006 #20


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    I understand but then it is wrong to say that the answer is in N/C!
    You are right that if you keep going this way, the x at the end will be in cm. But in order to get the right units I suggest to work in meters and to convert at the end only. If not, you should not say that you E's are in N/C.

    You are still mixing directions with magnitudes and vectors. First find the magnitudes and then determine the directions using the rule I gave you earlier.
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