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Homework Help: Electric Field Angle Problem

  1. May 6, 2010 #1
    1. The problem statement, all variables and given/known data

    The Figure below shows a plastic ring of radisu R = 50.0cm. Two small charged beads are on the ring. Bead 1 of charge +2.00 micro coulombs is fixed in the place at the left side of the ring. Bead 2 of charge +6.00 micro coulombs can be moved along the rignt. The two beads produce a net electric field of magnitude E at the center of the ring. At what (a) positive and (b) negative values of angle theta should bead 2 be positioned such that E = 2.00 x 10^5 N/C?

    I do not have the picture in electronic format so I will describe it:

    The figure is simply a circle on a xy coordinate system. Bead 1 is at the intersection of the x-axis and the circle on to the left of the origin. Bead 2 is on an arbitrary point on the circle in the second quadrant with a line of length R, from the origin, drawn to it. The angle, theta, is labeled as the angle between the line of length R and the x-axis.

    2. Relevant equations

    [tex]E = \frac{kQ}{R^2}[/tex]


    3. The attempt at a solution

    The E-field induced by each charged bead is found using the above equation. I can calculate the affect bead 1 (the fixed bead) will have.

    [tex]E_1 = \frac{kQ}{R^2}[/tex]

    Given the net E-field, I can determine the vector value for [tex]E_2[/tex] as [tex]E_1[/tex] and [tex]E_{net}[/tex] are in the same direction.

    [tex]E_2 = E_{net} - E_1[/tex]

    Using [tex]E_2[/tex] form above:

    [tex]E_2 = |E_2| (cos\vartheta + sin\vartheta)[/tex]

    At this point I get stuck... I'm not sure how to solve this for [tex]\vartheta[/tex]

    Did I do something wrong leading up to this... or is there some way to solve this that I'm not seeing?
     
    Last edited: May 6, 2010
  2. jcsd
  3. May 6, 2010 #2

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    Directionality is very important in this problem. I would re-write the above equation to something of the form,

    [tex] \vec E = \frac{kQ}{R^2} \hat r [/tex]

    where [itex] \hat r [/itex] is the unit direction vector from the charge to the test point.

    You need to be a little more specific here with your variables. There are two charges involved in the problem, so they should be labeled appropriately. Also, [itex] \vec E_1 [/itex] has its own direction.

    [tex]\vec E_1 = \frac{kQ_1}{R^2} \hat r_1 [/tex]

    In this case it's pretty easy to express [itex] \hat r_1 [/itex] in terms of [itex] \hat x [/itex] and [itex] \hat y [/itex]. (Here, I am labeling Cartesian coordinate unit vectors as [itex] \hat x [/itex], [itex] \hat y [/itex] and [itex] \hat z [/itex]. However, your textbook/coursework might use notation such as [tex] \hat i [/tex], [tex] \hat j [/tex] and [tex] \hat k [/tex], or maybe [itex] \hat a_x [/itex], [itex] \hat a_y [/itex] and [itex] \hat a_z [/itex]. Whatever the case, you should find it easy to express [itex] \hat r_1 [/itex] in terms of one of these Cartesian coordinate unit vectors.)

    Perhaps I am misunderstanding you here. [tex]\vec E_2[/tex], [tex] \vec E_1[/tex] and [tex]\vec E_{net}[/tex] are all in different directions.

    I'm not sure what to make of that. :uhh:

    Form the equation for [tex] \vec E_2 [/tex]. Its unit vector will point in the direction of [itex] \hat r_2 [/itex]. The trick here is then to express [tex] \hat r_2 [/itex] in terms of [itex] \hat x [/itex] and [itex] \hat y [/itex]. That's where the cosine and sines come in. Once you have the expression, you can (vector) sum it with [tex] \vec E_1 [/tex] to get [tex] \vec E_{net} [/tex]. :wink:

    [Edit: Okay, now I think I see what you mean by [itex]E_2 = |E_2| (cos \theta + sin\theta)[/itex]. But you need to include your Cartesian unit vectors in that equation (and there also might be '-' signs involved; remember the direction points from the charge Q2 to the center of the ring).]
     
    Last edited: May 6, 2010
  4. May 6, 2010 #3
    Thanks for your help!

    Wouldn't [tex]E_1[/tex] be in the [tex]-\hat i[/tex] direction?

    Concerning [tex]E_2[/tex],

    [tex]|E_2| (cos\vartheta -\hat i + sin\vartheta \hat j)[/tex]

    I would take these and add them (vector wise) for an equation with theta to solve for?
     
  5. May 6, 2010 #4

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    Oh, so close! http://www.websmileys.com/sm/sad/2.gif (that is, if I interpret your diagram description correctly, where Q1 is to the left of the origin.)

    I believe the direction is on the positive x-axis. Remember, the direction is from the charge to the test point (the test point is at the origin for this problem). So [tex] \hat r_1 = \hat i [/tex].

    The way I interpreted your description of the problem, is that the angle [itex] \theta [/itex] is with respect to the positive x-axis. In other words, if Q2 is in the second quadrant, [itex] \theta [/itex] is between 90o and 180o. Is that right?

    If so, try [tex]|E_2| (-cos\theta \hat i - sin\theta \hat j)[/tex]

    If I misinterpreted your description (such as the angle [itex] \theta [/itex] defined with respect to the negative x-axis, for example), just remember the direction is from the charge to the test location.

    [Edit: Yes, once you have your vector sum, you should solve for [itex] \theta [/itex].]
    [Another edit: Hint: The Pythagorean theorem may be involved in the process of solving for theta.]
     
    Last edited: May 6, 2010
  6. May 6, 2010 #5
    So I have:

    [tex]E_{NET} = \frac{k}{R^2} [(Q_1 - Q_2 cos \vartheta) \hat i - Q_2 sin \vartheta \hat j] [/tex]

    I'm not sure how the pythagorean theorem comes into play though... [tex]sin^2 \vartheta + cos^2 \vartheta = 1[/tex]
     
  7. May 6, 2010 #6

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    Looks good to me! :approve:

    [Edit: By the way, the above equation assumes that the angle [itex] \vartheta [/itex] starts from the positive x-axis. In other words, it means that if [itex] \vartheta = 0 [/itex], then Q2 is located at the point x = +0.5 m. This was my original assumption, based on your description of the figure. If my assumption is wrong, it will affect the above equation and the final answer to the problem!]

    The problem statement gave you the magnitude of [itex] E_{net} [/itex]. So now you need to find the magnitude of the above expression. In other words, you need to find magnitude of a vector, where the vector is composed of two perpendicular components. (Hint: you need to find the "length" of the hypotenuse. :wink: ["length" is not really a good word here, since your solving for electric field strength, not distance. But I'm having difficulty thinking of a better word.])
     
    Last edited: May 6, 2010
  8. May 6, 2010 #7
    So... square the i and j components, sum them, and take the square root?
     
  9. May 6, 2010 #8

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    There ya go! http://www.websmileys.com/sm/cool/049.gif

    (By the way though, I just want to make sure I'm not misleading you. The equation in your previous post assumes that if [itex] \vartheta = 0 [/itex] then Q2 lies on the point x = +50.0 cm, y = 0. That's the picture I have in my head based on the original description. But I'm a little unsure of exactly how the angle [itex] \vartheta [/itex] is defined.)
     
  10. May 7, 2010 #9
    I got it at this point. Thanks for all your help.

    And just for the record.. you were picturing it right.
     
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