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Electric Field

  1. May 23, 2006 #1
    I'm not sure if I've done this question correctly:

    An equilateral triangle with sides of 15.6 cm, and a charge of +2 uC at one vertex, and charges if -4uC each at the other two. Determine the electric field at the centre of the triangle.

    Basically what I've done is figure out the electric charge for each point, and then add them together. I'm not sure if this is even correct, and even if it is I'm not sure I understand why - I'm struggling a bit with the concepts.

    E1 = kQ1/r^2 = (9 x 10^9)(0.002)/(0.156)^2 = 7.39 x 10^8 N/C
    E2 = - 1.47 x 10^9 N/C
    E3 = -1.47 x 10^9 N/c

    Adding these three fields I got - 7.31 x 10^8 N/C as the electric field at the centre.

    Thanks
     
  2. jcsd
  3. May 23, 2006 #2

    Hootenanny

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    I don't think what you've done it quite correct, because you need the electric field at the centre of the triangle, therefore [itex]r\neq 0.156[/itex]. HINT: It may be easier if you combine the two negative charges into a single charge, i.e. find the centre of charge as you do for the centre of gravity. Then the problem just becomes a simple two charge particles problem.

    Do you follow?

    ~H
     
  4. May 23, 2006 #3
    Thanks, I realize now that r is definately not 0.156.

    I'm not sure I understand the centre of charge. Does this mean that at the middle the point charge would also be -4uC?
    I figured similar to centre of gravity, that 4(0.156)/2(4) = 0.078 m

    So at 0.078 m the charge would be -4uC

    Based on a right triangle, the total distance from the new point charge to the top of the triangle would be 0.135 m, and therefore the centre would be at 0.0675m.

    Now I just added the electric fields:
    E = (9 x10^9)(4 x 10-6)/(0.0675)^2 + (9 x10^9)(2 x 10-6)/(0.0675)^2
    E = 1.18 x 10^2 N/C

    Does this look correct now?
     
  5. May 24, 2006 #4

    Hootenanny

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    Ahh sorry, I think I've mislead you hear, centre of charge is not applicable. You must take the vector sum of the components, this page is useful http://hyperphysics.phy-astr.gsu.edu/hbase/electric/mulpoi.html#c2 . As the triangle is equilateral the distance (r) of the centroid of the triangle will be 0.078m. I realised my error this moring, but couldn't get to a computer to check it.

    Please accept my apology.

    ~H
     
  6. May 24, 2006 #5
    No problem I was thinking that it was all seeming a little too easy.
    So I've reworked the problem given the information provided above.

    The first thing is that I think that the centre is actually at 6.8cm, based on creating a right angled triangle and finding the height (13.5cm). Therefore the centre would be at half the height = 6.8cm

    So based on r = 0.068:
    E1 = kQ1/r^2 = (9 x 10^9)(0.002)/(0.068)^2 = 3.89 x 10^6 N/C
    E2 = 7.79 x 10^6 N/C
    E3 = 7.79 x 10^6 N/C

    Then I found E1x, E2x, and E3x using Ecos30
    E1x = 3.37x10^6, E2x = 6.75 x 10^6, E3x = 6.75 x 10^6
    Adding these together Ex = 1.69 x 10^7

    Then E1y, E2y, E3y using Esin30
    E1y = 1.95x10^6, E2y = 3.89x 10^6, E3y = 3.89x 10^6
    Adding these together Ey = 9.73 x 10^6

    Therefore E = sqrt[(1.69 x 10^7)^2 + (9.73 x 10^6)^2]
    E = 1.95 x 10^7

    First does this like I'm doing it right now? And secondly I have not taken the negative charges into consideration and am wondering if I need to do so?

    Thanks
     
    Last edited: May 24, 2006
  7. May 25, 2006 #6

    Hootenanny

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    Your method looks ok, however as you say, you haven't taken into account the polarity of the charges. Note that the electric field of a single point charge is given by;

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

    All you need to do is make sure you substitute the value in with the positive or negative sign.. For example for one of the negative charges you would use [itex]Q = -4\times 10^{-6}[/itex]. Just one thing I need to check when you write "uC" do you mean "milli" or "micro" coulombs? Milli (m) = [itex]10^{-3}[/itex] and Micro ([itex]\mu[/itex]) = [itex]10^{-6}[/itex].

    ~H
     
  8. May 25, 2006 #7
    I did mess up the units.

    So now I have:
    E1 = kQ1/r^2 = (9 x 10^9)(2 x 10^-6)/(0.068)^2 =1.22 x 10^3 N/C
    E2 = -2.45 x 10^3 N/C
    E3 = -2.45 x 10^3 N/C

    Then
    E1x = 1.06 x 10^3, E2x = -2.12 x 10^3, E3x = -2.12 x 10^3
    Adding these together Ex = -3.18 x 10^3

    Then
    E1y = 6.1 x 10^2, E2y = -1.23 x 10^3, E3y = -1.23 x 10^3
    Adding these together Ey = -1.85 x 10^3

    Therefore E = sqrt[(-3.18 x 10^3)^2 + (-1.85 x 10^3)^2]
    E = 3.7 x 10^3

    Does this look better now with the charges put in?

    Thanks so much for the help
     
  9. May 25, 2006 #8

    Hootenanny

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    Yeah, that looks good to me (I assume that you've punched the numbers into your calculator correctly:wink:). Don't forget the units of electric field strength, most tutors deduct marks for incorrect or missing units!

    Again I apologise for the error at the start (I don't know what I was thinking!)

    ~H
     
  10. May 25, 2006 #9
    No problem, thank you for the much needed help!
     
  11. Jul 2, 2007 #10
    Check you math here. I calculated 3.89 x 10^6 N/C
     
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