1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    User Avatar
    Homework Helper
    Gold Member

    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


    User Avatar
    Homework Helper
    Gold Member

    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


    User Avatar
    Homework Helper
    Gold Member

    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


    User Avatar
    Homework Helper
    Gold Member

    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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook