Electric Fields-Calculating the total force on a charge

  • Thread starter rico22
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  • #1
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Homework Statement


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




Homework Equations



λ=Q/l
l=pi/R
∫dE=∫(k ldql/r2

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.
 

Answers and Replies

  • #2
tms
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Don't forget that [itex]\lambda[/itex] is a function of [itex]\vartheta[/itex]: [itex]\lambda(\vartheta)[/itex].
 
  • #3
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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)?
 
  • #4
tms
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It would probably be easiest to do each component separately.
 
  • #5
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you mean r[∫λ∫cos(θ)]?
 
  • #6
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I mean am I even on the right path with my first attempt? Is this the only thing I did wrong?
 
  • #7
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anybody?
 
  • #8
haruspex
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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)?
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?
 
  • #9
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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.
 
  • #10
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as far as the direction of the field goes I would guess that its in the downward (negative y) direction???
 
  • #11
haruspex
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as far as the direction of the field goes I would guess that its in the downward (negative y) direction???
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.
 
  • #12
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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?
 
  • #13
haruspex
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so in this case ∫λ0cos(θ) would not equal λ0sin(θ) from -pi/2 to +pi/2? correct?

Why not?
 

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