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Homework Help: Electric Force, linear charge density

  1. May 26, 2006 #1
    First I want to say thanks in advance - I found this site through Google and am thrilled! All I have left to graduate is Physics II (Electromagnetism & Waves), so I'll finally be walking on August 5. IF I can pass this class, that is.

    I'm a non-traditional student, and it's been about 10 years since I had the first physics, or calculus. So I'm struggling...

    I ran across this discussion (https://www.physicsforums.com/showthread.php?t=92557) thanks to Google, which lists my question almost to the T, but I'm still lost - I can't figure out where to go from here. I feel like our professor hasn't shown us nearly enough examples - he's spent most of the lecture time talking about theory, which makes sense to me until he throws math at me. Then I'm lost!

    Remember I haven't had calculus in years and years, so bear with me....

    A line of positive charge is formed into a semicircle radius R=0.6m.
    The charge per unit length is given by [tex]\lambda=\lambda_{0}cos\theta[/tex].
    The total charge on the semicircle is 50[tex]\mu C[/tex] Calculate the total force on a charge of 8[tex]\mu C[/tex] placed at the centre of the curvature.
    The semicircle is like the top half of a circle with center as the origin. The angle [tex]\theta[/tex] is measured from the y axis.
    (thank God for the quote button while I figure this latex stuff out!!)

    Ok, I have no idea if this is right or not, but what I came up with is (I guess you can't specify limits of integration w/ latex?)
    [tex]F= \frac {kq_{0}} {r^2} \int cos\theta d\theta [/tex]

    uh.... I think? integrate that from 0 to [tex]\pi[/tex]? but the integral of cos is -sin, right? and sin 0= sin[tex]\pi[/tex]= 0. right?

    and what the heck does the total charge of 50[tex]\mu C[/tex] have to do with anything?

    I'm so lost....
    Last edited: May 26, 2006
  2. jcsd
  3. May 26, 2006 #2

    Andrew Mason

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    50[itex]\mu C[/itex] is the sum of all the incremental charges, dq from one end to the other:

    [tex]dq = \lambda_0 \cos\theta rd\theta[/tex]

    Integrate that from 0 to [itex]\pi[/itex] and equate to the total charge to get the value for [itex]\lambda_0[/itex].

    [tex]dq = \lambda_0 \cos\theta rd\theta[/tex]

    [tex]dF= \frac{kq_{0}dq}{r^2} [/tex]

    Work out the x and y components of dF and then integrate:

    [tex]F_y = \int_0^\pi dF_y[/tex]

    [tex]F_x = \int_0^\pi dF_x[/tex]

    Just work out the value of those integral and plug in values to get the total force. Since it appears to be symmetrical about the y axis, the net force should be along the y axis.

    Last edited: May 26, 2006
  4. May 27, 2006 #3
    Note the [itex]cos\theta[/itex] behaviour here ..... half of the semicircle from 0 to [tex]\frac{\pi}{2}[/tex] will be positively charged , thus force vector acting away from that part of the region ...and from [tex]\frac{/pi}{2}[/tex] to [tex]/pi[/tex] , will be negative , integrate for both halfves seperately , and for integration for each part , seperate the dF component in x and y components and then integrate , after you are done for both regions , add the x and y vectors of force. You will notice the net force points towards -vely charged region...semicircle acts like a dipole to some extent.
  5. Jan 27, 2008 #4
    I am having some trouble setting up the x and y components of dF. It has been awhile since I had mechanics and it appears to me that the x and y components of the resultant dF from 2 lines going through the center of the semicircle is not going to have the equivalent theta as its angle for the right triangle composed of its components. This can be seen in the pic I uploaded (if you can't see it very well just zoom in or load in paint and zoom, the quality won't decay because it is so low res already).

    I am kinda confused as to how to setup these components with the equation I already have.

    hopefully i got the latex right...

    [tex]dF = \frac{kq_{0}dq\lambda_0 \cos\theta} {r} d\theta [/tex]

    the r's cancel leaving me one in the top
    now how do i find this beasts components so i can integrate them?

    I used my equation and multiplied times cos theta to make it the Y component and then integrated. I got exactly twice what the answer is ... wonder what I did wrong. I realized that cos theta time dF would the Y component due to geometry and the X component is null anyway. I have no clue how to find Y and X if this weren't true : /

    sooo i know now how to compute the answer but I still need help with why I am having to halve it to find the correct one...
    also my [tex]\lambda_0 = \fraq{Q}{r}[/tex] and where I found Q by integrating one side of the semi-circ... oh **** that is why cause I need to acount for the other side which will cut this in half because Q should be [tex]Q = 2\lambda_0 r [/tex]
    ya so i guess i figured it out by talking to myself in this post lol

    ya .... soo.... i guess my only question is about how you would normally figure out the components dFx and dFY in this case if you already have dF

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    Last edited: Jan 27, 2008
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