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Electric Field, solving for distance

  1. Jan 18, 2008 #1
    [SOLVED] Electric Field, solving for distance

    1. The problem statement, all variables and given/known data
    A 18.8 nC charge is located at position ( x,y ) = (1.0cm, 2.0cm). At what (x, y) position(s) is the electric field
    a) -225,000i^ N/C (Where i^ is "i" with a hat on it... easier to explain this than look up the text code)
    b) (161,000i^ + 80,500j^) N/C
    c) (21,600i^ - 28,800j^) N/C

    2. Relevant equations
    E = kQ/r^2
    Aka, r = Sqrt(169.2 / E)

    3. The attempt at a solution
    a) Plugged in E into r = Sqrt(169.2 / E). I then factored in the fact that the charge is not at the origin. Got a final answer of -1.74cm. This is correct.

    b) Tried the same thing here. There are two answers I need, one for the X position, one for the Y. I tried ignoring the Y, and just solving the X position using 161,000i^. Factored in the fact that the charge is not at the origin, got +4.24cm. This is incorrect.

    I see that my reasoning works in 1d, but once we involve the 2nd dimension, something is screwing me up.

    What am I not taking into account?
  2. jcsd
  3. Jan 18, 2008 #2
    hmm.. one method i can guide you through is:

    as you know, the electric field for a point charge extends radially outwards. So, if i give you a particular magnitude of electric field, say [itex]\lambda[/itex] [remember, not the direction, but only the magnitude], then all the points where the electric field is [itex]\lambda[/itex] will lie on a circle. The radius of the circle is what we are interested in. For your given [itex]E = 161,000\hat{i} + 80,500\hat{j}[/itex], find [itex]|E|[/itex]. Then using the formula: [tex]E = \frac{kq}{r^2}[/tex], find 'r'.

    Once you have found the radius of the circle, can u find the point on it having a position vector which has a similarity to the vector field [HINT: I'm thinking direction ratios]

    sorry.. can't post the whole solution.. have been warned by the pf staff for doing that.. and well.. i agree with them on this matter...
  4. Jan 18, 2008 #3
    Ahhh, yeah, I see what you're getting at. Thanks ^^;
  5. Jan 18, 2008 #4
    Sigh. It worked for the 2nd question, and now the 3rd (Which is just like the 2nd) doesn't seem to work. I've double-checked my numbers, they all should work.

    q = 18.8*10^-9
    k = 9*10^9
    E = (21600i^ - 28,800j^)
    Charge coordinates = (1cm, 2cm)

    E = kQ/r^2
    Therefore r = Sqrt(kQ/E)
    kQ = 9 * 18.8 = 169.2 (The 10^9 and 10^-9 cancel out)

    r = Sqrt(169.2/E)

    The value for E is found by using the line-distance formula:
    E = Sqrt(21600^2 + 28800^2)
    E = Sqrt(466560000 + 829440000)
    E = Sqrt(1296000000)
    E = 36000

    r = Sqrt(169.2/E)
    r = Sqrt(169.2/36000)
    r = Sqrt(0.0047)
    r = 0.068556546004010441249358714490848 m
    r = 6.8556546004010441249358714490848 cm (So we're working with common units)

    The x value should be = r*cos(arctan(21600 / -28800)) + 1
    x = 6.8556546004010441249358714490848 * cos(arctan(21600 / -28800)) + 1
    x = 6.48455cm
    Radius of the circle, multiply by cos of the angle (Angle is calculated by doing arctan of the opposite over the adjacent), then adding an offset of 1, since this is relative to the origin, not the charge emitting it.
    The y value should be = r*sin(arctan(21600 / 28800)) + 2
    Since my x is wrong, not computing the y yet.

    I think it might have something to do with which quadrant this is in? What do you guys think?
  6. Jan 18, 2008 #5
    Took quadrants into account, and got the right answer.
    Thanks for the circle tip!
  7. Jan 21, 2008 #6
    How does different quadrants affect the answer. So if its in the 4th quadrant what would you change from your steps to solve this problem. I am trying to figure out how to do (c) but i cant seem to find a way..... please clarify for me the diffrence between quadrants!
    thanks in advance.
  8. Jan 21, 2008 #7
    "The x value should be = r*cos(arctan(21600 / -28800)) + 1"

    Arctan(Opposite/Adjacent) gives the angle
    Then take the cos of that angle.

    arctan(21600 / 28800) = 36.8699°
    However, picturing a circle, this would imply the first quadrant (Positive X value, positive Y value).

    Yet, our field is in the 4th quadrant (Positive X, negative Y), therefore you need to add 270° to your arctan to change it from first to fourth quadrants.

    36.8699° + 270° = 306.8699°.

    Then take the cos, multiply by r, etc, etc, etc, to get your final answer.
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