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Electric potential - Is the teacher wrong?

  1. Feb 23, 2008 #1

    tony873004

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    electric potential -- Is the teacher wrong?

    1. The problem statement, all variables and given/known data
    This is from our practice midterm test:

    A point charge of 8.0 [tex]\mu C[/tex] is placed at the origin of a coordinate system and a charge of -7.0 [tex]\mu C[/tex] is placed at (0 m, 0.25 m). Find where on the y-axis the potential is zero.


    2. Relevant equations
    V=kQ/y


    3. The attempt at a solution
    [tex]
    \begin{array}{l}
    d_1 = y \\
    d_2 = y - 0.25 \\
    \\
    \frac{{Q_1 }}{y} = \frac{{Q_2 }}{{y - 0.25}}\,\,\,\, \Rightarrow \,\,\,\,\frac{{Q_1 }}{y} = \frac{{Q_2 }}{{y - 0.25}}\,\,\,\, \Rightarrow \,\,\,\, \\
    \\
    \,Q_1 \left( {y - 0.25} \right) = yQ_2 \,\,\,\, \Rightarrow \,\,\,\,Q_1 y - 0.25Q_1 = yQ_2 \,\,\,\,\, \Rightarrow \,\,\,\, \\
    \\
    - 0.25Q_1 = yQ_2 - Q_1 y\,\,\,\, \Rightarrow \,\,\,\, - 0.25Q_1 = y\left( {Q_2 - Q_1 } \right)\,\,\,\, \Rightarrow \\
    \\
    y = \frac{{ - 0.25Q_1 }}{{\left( {Q_2 - Q_1 } \right)}} = \frac{{ - 0.25\left( {8\mu C} \right)}}{{ - 7\mu C - 8\mu C}} = \frac{{ - 2\mu C}}{{ - 15\mu C}} = \frac{2}{{15}}{\rm{m}} \\
    \end{array}
    [/tex]

    Here's the problem. In class, he said it would boil down to a quadratic, and we should get two answers. On the worksheet, he gives two answers of 2/15, like I got, and 2. But I never got a quadratic. Furthermore, I played around with the online applet for 2 opposite charges, and I don't see the equipotential line for 0 crossing the y-axis more than once. Here's a screen shot: [​IMG]

    What am I missing here?
     
    Last edited: Feb 23, 2008
  2. jcsd
  3. Feb 23, 2008 #2

    tony873004

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    I'm guessing the teacher is right. In the applet, my charges are equal. It doesn't let me control that. But in the example they are not quite equal, so perpahs that zero line does wrap around. Also, simply plugging (2/15) and (2) into the 8/y = -7/(y-0.25) gives me equal but opposite answers in both cases. But I still don't get a quadratic. My method implies that there are not two answers.
     
  4. Feb 23, 2008 #3
    Yah I think you did something wrong, let me make sure I understand this right...

    Q2 is located at the point (m, .25m)?

    So you're looking for the charge's distance to the y axis, which for Q1 is simply enough y since it's on the origin, but the distance from Q2 to y is gonna be sqrt[m^2+(y-.25m^2)^2] where y is whatever point on the y axis

    As for why there's two answers your picture from the applet, while sweet, is implying both charges are sitting on the y axis, which barring typo or me reading it wrong is not the case. Especially if the offset charge is smaller, that "zero line" is gonna curve back and touch the y axis twice
     
  5. Feb 23, 2008 #4

    tony873004

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    sorry, its a typo.

    charge 2 is at (0 meters, 0.25 meters). Both charges are on the y-axis. The point(s) of interest are also on the y-axis.

    Thinking about it, I should probably be doing kq1/y + kq2/(y-0.25) = 0 rather than kq1/y = kq2/(y-0.25). At least that explains the "equal but opposite" dillemma. But I'm still not getting a quadratic.
     
    Last edited: Feb 23, 2008
  6. Feb 23, 2008 #5

    D H

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    Think about it this way: as [itex]y\to\pm\infty[/itex] it begins to look more and more like a charge of [itex]1\mu C[/itex] at the origin. Now envision a test point moving from y = plus infinity toward y = minus infinity. At some point the test point will see the closer but smaller negative charge as exactly balancing out the positive charge at the origin. There are obviously poles at 0.25 m and the origin. Somewhere between these poles the negative charge at 0.25 m and the positive charge at the origin will balance.

    In general, the equipotential surfaces for this charge distribution will not be a "quadratic". However, the special case of zero potential is "quadratic" (I put "quadratic" in quotes because the correct term is quadric, not quadratic).
     
  7. Feb 24, 2008 #6
    Yah, note that now with the right form of the equation (V1+V2=0) I'm gonna bet 2/15*m, when plugged in doesn't work. Where as with the other way 2m didn't work. I believe that makes sense, explain!
     
  8. Feb 24, 2008 #7

    D H

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    The potential is zero at 2/15 meters and 2 meters. The potential along the y-axis is quadratic in y.
     
  9. Feb 24, 2008 #8

    tony873004

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    I believe you because when I plug them in to verify the answers, they work. But how do you get that? I don't get a y^2, I only get a y.
     
  10. Feb 24, 2008 #9

    tony873004

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    [tex]\begin{array}{l}
    \frac{{kq_1 }}{y} + \frac{{kq_2 }}{{y - 0.25}} = 0 \\
    \\
    \frac{{kq_1 \left( {y - 0.25} \right) + kq_2 y}}{{y\left( {y - 0.25} \right)}} = 0 \\
    \\
    kq_1 \left( {y - 0.25} \right) + kq_2 y = 0 \\
    \\
    kq_1 y - 0.25kq_1 + kq_2 y = 0 \\
    \\
    kq_1 y + kq_2 y = - 0.25kq_1 \\
    \\
    y\left( {kq_1 + kq_2 } \right) = - 0.25kq_1 \\
    \\
    y = \frac{{ - 0.25kq_1 }}{{kq_1 + kq_2 }} \\
    \\
    y = \frac{{ - 0.25q_1 }}{{q_1 + q_2 }} \\
    \end{array}[/tex]

    There's no y2, hence only 1 value for y. What am I doing wrong?
     
  11. Feb 24, 2008 #10

    D H

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    You are dividing by y and y-0.25. You should be dividing by the distance from the point charges. For example, what is the potential at some point away from the y-axis? With point charges [itex]q_1[/itex] and [itex]q_2[/itex] at [itex]\mathbf r_1[/itex] and [itex]\mathbf r_2[/itex], the potential is

    [tex]
    U(\mathbf r) =
    \frac 1{4\pi\epsilon_0}
    \left(
    \frac {q_1}{||\mathbf r - \mathbf r_1||} +
    \frac {q_2}{||\mathbf r - \mathbf r_2||}
    \right)
    [/tex]

    The surface on which the potential is zero is thusly

    [tex]
    \frac {q_1}{||\mathbf r - \mathbf r_1||} +
    \frac {q_2}{||\mathbf r - \mathbf r_2||} = 0
    [/tex]

    or

    [tex]
    q_1||\mathbf r - \mathbf r_2|| = -q_2||\mathbf r - \mathbf r_1||
    [/tex]

    Setting [itex]q_1=8\mu C[/itex], [itex]q_2=-7\mu C[/itex], [itex]\mathbf r_1 = \mathbf 0[/itex], and [itex]\mathbf r_2=0.25\,\text{m}\,\hat y[/itex],

    [tex]
    8||\mathbf r-0.25\,\text{m}\,\hat y|| = 7 ||\mathbf r||
    [/tex]

    Squaring to get rid of those vector magnitudes,

    [tex]
    64 (r^2 - 0.5\,\text{m}\,\mathbf r\cdot\hat y + \frac 1 {16} \,\text{m}^2)
    = 49 r^2
    [/tex]

    For a point on the y-axis ([itex]\mathbf r = y\hat y[/itex]) this becomes

    [tex]
    64 (y^2 - 0.5\,\text{m}\,y + \frac 1 {16} \,\text{m}^2) = 49 y^2
    [/tex]

    which simplifies to

    [tex]
    15 y^2 - 32\,\text{m}\,y + 4 \,\text{m}^2 = 0
    [/tex]

    This quadratic has solutions

    [tex]y=\frac{32\pm28}{30} = \frac 2 {15}, 2[/tex]
     
  12. Feb 24, 2008 #11

    tony873004

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    Thanks for that detailed answer! I follow the 1st part, up to where you said "squaring". I'll look at it tommorow. My brain is fried now. Thanks, DH
     
  13. Feb 24, 2008 #12
    The part where you squared r, did you square the vector of y-cap too?

    I did it in a weird way, but I got 2/15 and 2. Although I got the answers, I'm not entirely sure it was correct.

    I set the 0,0 coor as y and 0, 0.25 as y+0.25
    So 8/y + -7/(y+0.25) = 0
    8/y = -[-7/(y+0.25)]
    8/y = 7/(y+0.25)
    y = -2

    I also tried 8/y - [-7/(y+0.25)]=0
    8/y = [-7/(y+0.25)]
    y = -2/15

    Can you please explain why the way I used only works when d1=y and d2 = y - 0.25?
    And is the way I did it valid?
     
  14. Feb 24, 2008 #13

    D H

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    Answering both of these questions at once, the Euclidean norm of some vector [itex]\mathbf x[/itex] is, by definition, the number [itex]r[/itex] such that [itex]r*r = \mathbf x \cdot \mathbf x[/itex]. So what if [itex]\mathbf x = \mathbf r - 1/4\,\text{m}\,\hat y[/itex]?

    [tex]
    \begin{aligned}
    ||\mathbf r - \frac 1 4 \,\text{m}\,\hat y||^2 &=
    (\mathbf r - \frac 1 4 \,\text{m}\,\hat y) \cdot
    (\mathbf r - \frac 1 4 \,\text{m}\,\hat y) \\[4pt]
    &= r^2 - \frac 1 2 \,\text{m}\,\mathbf r\cdot\hat y + \frac 1 {16} \,\text{m}^2
    \end{aligned}
    [/tex]

    Look at your own post again. You did not get 2/15 and 2. You got -2/15 and -2. So what did you do wrong? The potential due to the charge at the origin is not 8/y. What is the potential at y=-1?

    Ignoring off axis points, one can do something similar to what I did above but using absolute values. The potential due to the charge at the origin is 8/|y|, not 8/y. The same goes for the charge at 0.25 meters.

    [tex]
    \begin{aligned}
    & \frac 8 {|y|} + \frac {-7} {|y-0.25|} = 0 \\
    & 8 |y-0.25| = 7 |y| \\
    & 64 (y-0.5y+0.0625) = 49 y^2 \\
    & 15 y^2 - 32y + 4 = 0
    \end{aligned}
    [/tex]

    Which is of course the same quadratic as before (except here I was sloppy with the units).
     
  15. Feb 24, 2008 #14
    I understand the absolute values, but I'm confused about why the second d is y-0.25 instead of y+0.25
    I would assume |y| is the first d (the lower coordinate) and |y+0.25| is the second d (the higher coor at 0,0.25) But if you use |y-0.25| wouldn't this coordinate be lower than |y|?
     
  16. Feb 24, 2008 #15

    D H

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    You use y-0.25 because the second charge is 0.25 meters above the origin. Think about it. Which form describes the distance when the test point y is also located 0.25 meters above the origin: |y-0.25| or |y+0.25| ?
     
  17. Feb 24, 2008 #16
    Oh I see.. I always thought y was used to represent point 1. Didn't know it was used as the test charge.
     
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