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Finding electric field of charges in a rectangle

  1. Apr 23, 2014 #1
    1. The problem statement, all variables and given/known data
    Find the electric field of a charge at a corner if there are 4 equal charges at the corner of a rectangle with sides 40 cm and 20 cm. The charges are 2.5 x 10^3 C.


    2. Relevant equations
    E = (kq)/r^2


    3. The attempt at a solution
    I labelled the top left charge q1, the top right charge q2, the bottom left charge q3, and the bottom right charge q4. The corner I picked to find the electric field was the top right corner, at q2. So what I'm going to do is find all the charges acting horizontally and vertically, then I apply pythagorean's theorem to find the net charge. Is that right so far?

    So I took the charge of q1 acting on the top right corner, the charge of q4 acting upwards on the top right corner, and the charge of q3 on the top right corner. I'll put in some numbers after I know I'm on the right track.
     
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  3. Apr 23, 2014 #2

    mfb

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    What do you mean with "net charge"?

    You have to add the contributions, right.
     
  4. Apr 23, 2014 #3
    I'll have all the charges acting in the x component, and all the charges in the y component, so by doing sqrt(x^2+y^2), I'll get the net charge of the electric field acting on that corner.
     
  5. Apr 23, 2014 #4

    mfb

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    What does "doing sqrt(x^2+y^2)" mean? "Do 3.4"?
    There is no need to calculate anything that could be described as net charge.
    One charge acts in both components at the same time and you have to consider this.
     
  6. Apr 23, 2014 #5
    There is one charge that acts in both components, I have broken it down into two separate components, one x and y. The 2nd charge acts on horizontally, and the 3rd charge acts on vertically. Then, I added all the x and y components together. Don't i need sqrt(x^2+y^2) to add up the x and y components
     
  7. Apr 23, 2014 #6
    So E(q1) = k(q1)/r^2, putting all those numbers gives me 1.4 x 10^8 N/C (right)
    E(q4), which is the charge on the bottom right, equals 5.62 x 10^ 8 (upwards)
    E(q3) is the charge that is diagonally away from the electric field that we want to find, so we need to break into two components, the x component is E(q3x) = (kq3cos(theta))/r^2, where r^2 is the hypotenuse sqrt(5)/5, and E(q3y), the y component, is (kq3sin(theta))/r^2
     
  8. Apr 24, 2014 #7

    mfb

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    Speak in full physical statements, please. There are many wrong ways to use that expression.

    The length is not sqrt(5)/5 (that is just one part of the expression), but apart from that it looks good so far.
     
  9. Apr 24, 2014 #8
    Converting cm to meters, I have 0.4m by 0.2m. So by using Pythagorean theorem, I'll get the length of the diagonal across the rectangle, which i calculated to be sqrt(5)/5. Is that correct?

    I'm using Pythagorean theorem to find the net electric field because for the charge at the corner, it will experience an electric field acting horizontally, and an electric field acting vertically. Thus, by adding the components, I can obtain the net electric field acting on the charge at the corner. Is this use of the theorem correct?

    My friend did the exact same problem and we both got different answers. I did it this way, while he did it by calculating the force acting on the charge at the corner, and then at the end, divide by energy? to get the electric field. He broke the problem down into components too, so I'm not sure why we got different answers if he calculated force first.

    If the process is correct, I will proceed with the calculations.(and I will try to use LaTex)
    Thanks!
     
    Last edited: Apr 25, 2014
  10. Apr 25, 2014 #9
    I would edit my post, but it won't let me. My friend divided by charge at the end (2.5 x 10^4), not energy.

    By the way, someone copied my exact question and posted it on some random site....
    http://physicsinventions.com/index.php/finding-electric-field-of-charges-in-a-rectangle/ [Broken]

    What's up with that? Weird.
     
    Last edited by a moderator: May 6, 2017
  11. Apr 25, 2014 #10

    SammyS

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    There is a time limit for editing a post. It seems to me it's something like 700 minutes - that's 11 hours and 40 minutes.


    Calculating the force, then dividing by the appropriate charge should give the same answer as calculating the field directly.
     
    Last edited by a moderator: May 6, 2017
  12. Apr 26, 2014 #11

    mfb

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    sqrt(5)/5 is a real number, it is not a length.

    Could be, your description is not detailed enough to tell.

    That is possible as well. You just have one more factor to care of, it drops out afterwards anyway.

    Some pages try to steal content, our administrators will have a look at it and try to get that copy removed.
     
  13. Apr 26, 2014 #12
    So are you looking for a more detailed explanation of my use of the pythagorean? I'm not sure if I can do any better. To me, sqrt(5)/5 is the magnitude of the x and y component vectors, so a length, but I could use tanΘ to determine the angle it is from the horizontal. Together, I could determine the net direction and magnitude/length of the electric field, which is a vector. I could be wrong, but it's how I think about it.
     
  14. Apr 26, 2014 #13

    mfb

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    A length has units of length. sqrt(5)/5 does not have that.
    In the same way "1 day" and "100 kg" are not lengths, sqrt(5)/5 is not a length as well. sqrt(5)/5 meters? sqrt(5)/5 centimeters? sqrt(5)/5 something else?

    You could do it in the same way you plan to hand in your homework.
     
  15. Apr 26, 2014 #14
    In that case, I meant meters.
     
    Last edited: Apr 26, 2014
  16. Apr 26, 2014 #15
    q = q1 = q2 = q3 = q4

    [itex]\varepsilon_{q1} = \frac{kq}{r^2} = \frac{8.99(10^9)(2.5* 10^{-3})}{(0.4)^2}=1.4 * 10^8 N/C [/itex] (right)

    [itex]\varepsilon_{q4} = \frac{kq}{r^2} = \frac{8.99(10^9)(2.5* 10^{-3})}{(0.2)^2}=5.62 * 10^8 N/C [/itex] (up)

    [itex]sin \theta = \frac{100}{\sqrt{5}}, cos\theta = \frac{200}{\sqrt{5})} [/itex]

    To split the 3rd charge into components:

    [itex]\varepsilon_{q3x} = \frac{kq cos\theta}{r^2} = \frac{8.99(10^9)(2.5* 10^{-3}(\frac{200}{\sqrt{5}})}{(\frac{\sqrt{5}}{5})^2}= 1.00 * 10^{10} N/C [/itex] (right)

    [itex]\varepsilon_{q3y} = \frac{kq sin\theta}{r^2} = \frac{8.99(10^9)(2.5* 10^{-3}(\frac{100}{\sqrt{5}})}{(\frac{\sqrt{5}}{5})^2}= 5 * 10^9 N/C [/itex] (up)

    [itex]\varepsilon_{net} = \sqrt{[1.00(10^{10}) + 1.4 * 10^8)]^2 + [(5.62 (10^8) + 5 (10^9)]^2} = 1.16 (10^{10}) [/itex] -> magnitude of electric field

    I've never used latex in my life, so I hope I'm doing it right.
     
  17. Apr 26, 2014 #16

    SammyS

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    The latex looks very good !

    As regards getting the components for ## \vec\varepsilon_{q3}\ ##, it appears that you are not using the full distance between the charges. You should only multiply the magnitude of this vector by a sine or cosine, as appropriate.


    (This could have been cleared up much sooner if you had included these details in earlier posts.)
     
    Last edited: Apr 26, 2014
  18. Apr 26, 2014 #17
    That's true, but I really didn't want to learn LaTex then, haha. Because it is the y component, shouldn't I multiply by the sine? (just as I used cosine for the x component)
     
  19. Apr 26, 2014 #18

    SammyS

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    Yes. I should NOT have had the y subscript at all. I should have had, ##\vec\varepsilon_{q3}\ ## .

    I'll edit that.
     
  20. Apr 26, 2014 #19
    Does that mean I also have to multiply the lengths [itex] (\frac{\sqrt{5}}{5})[/itex] by the appropriate sine or cosine as well?
     
  21. Apr 26, 2014 #20

    SammyS

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    No.

    Multiply the magnitude of ##\ \vec\varepsilon_{q3} \ ## by the appropriate sine or cosine.
     
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