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Electric Fields-Calculating the total force on a charge

  1. Feb 8, 2013 #1
    1. The problem statement, all variables and given/known data
    A line of positive charge is formed into a semicircle of radius R = 40.0 cm. The charge per unit length along the semicircle is described by the expression λ = λ0 cos(θ). The total charge on the semicircle is 15.0 µC. The semi circle starts off at 0 and finishes at pi going through quadrants I and II. Calculate the magnitude of the total force on a charge of 2.00 µC placed at the center of curvature. θ is to be taken to start off from pi/2 thus creating a triangle with R that cos(θ)= y/R and sin(θ)=x/R




    2. Relevant equations

    λ=Q/l
    l=pi/R
    ∫dE=∫(k ldql/r2

    3. The attempt at a solution
    I replaced ldql with λRdθ. then I found lambda to be 1.194x10-5 since I know the length of the semicircle and total charge Q. I plugged this into the integral to find E(electric field) and got Ex=kλ/R∫sin(θ) ⇔ kλ/R (-cosθ). this should give me the magnitude of the electric field. then I multiplied this time the charge of 2.00μC placed at the center of the curvature. Where am I going wrong? Any response would be greatly appreciated.
     
  2. jcsd
  3. Feb 8, 2013 #2

    tms

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    Don't forget that [itex]\lambda[/itex] is a function of [itex]\vartheta[/itex]: [itex]\lambda(\vartheta)[/itex].
     
  4. Feb 9, 2013 #3
    so since I have Q can I just integrate dq=λds=λrdθ?

    so Q=∫rλdθ=r∫λdθ??? but this is where I get confused should I integrate λcos(θ)?
     
  5. Feb 9, 2013 #4

    tms

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    It would probably be easiest to do each component separately.
     
  6. Feb 9, 2013 #5
    you mean r[∫λ∫cos(θ)]?
     
  7. Feb 9, 2013 #6
    I mean am I even on the right path with my first attempt? Is this the only thing I did wrong?
     
  8. Feb 9, 2013 #7
    anybody?
     
  9. Feb 9, 2013 #8

    haruspex

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    Yes. That's how to find λ0.
    I found the statement of the problem a bit confusing, but I think it is measuring theta as the angle from the positive Y axis, clockwise being positive for the angle. So theta runs from -pi/2 at (-R, 0) through 0 at (0, R) to pi/2 at (R, 0). Is that your understanding?
    On that basis, can you immediately deduce what direction the field must be in at the origin?
     
  10. Feb 9, 2013 #9
    thank you for the reply...yes I know that the limits of integration would be from -pi/2 to +pi/2... im just confused as to whether im in the right track as far as the integration goes being that λ in this problem is a function of θ.
     
  11. Feb 9, 2013 #10
    as far as the direction of the field goes I would guess that its in the downward (negative y) direction???
     
  12. Feb 10, 2013 #11

    haruspex

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    Yes, which means you only need to include the component in that direction in the integrand.
    But first, do the integration of the charge to find λ0.
     
  13. Feb 10, 2013 #12
    But first, do the integration of the charge to find λ0

    Well that's what im having problems with being that λ is a function of θ, usually I would just take it out of the integrand...so in this case ∫λ0cos(θ) would not equal λ0sin(θ) from -pi/2 to +pi/2? correct?
     
  14. Feb 10, 2013 #13

    haruspex

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    Why not?
     
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