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modphysnoob

  • one year ago

I am doing practice problem and have no solution to check my work. so I would post my work and please tell me whether or not it is correct

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  1. modphysnoob
    • one year ago
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    3.51) a) Since the triangle is in x -y plane , its vector would be in z direction. but E is only in x and y direction , hence it is 0 @JamesJ

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  2. JamesJ
    • one year ago
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    No, what you just wrote for part a is wrong. You are evaluating a line integral.

  3. JamesJ
    • one year ago
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    The first thing I would check is if this E field is conservative, i.e., curl-free. If so, then the line integral over any closed loop must be zero. If E is not conservative, then you will have to calculate the three pieces explicitly.

  4. modphysnoob
    • one year ago
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    it does seems to be conservative

  5. modphysnoob
    • one year ago
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    doesn't

  6. modphysnoob
    • one year ago
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    x =/2x

  7. JamesJ
    • one year ago
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    Right, it has non-zero curl. So evaluate the line integral along each of the three edges.

  8. modphysnoob
    • one year ago
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    ok , so we start with horitozotal line along x axis Integrate[E dl] E= {xy,-x^2-2y^2} dl that part is {1 , 0} Int[ x*y dx ] from 0 to 2

  9. JamesJ
    • one year ago
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    ...where y = 0. Hence the integral = 0

  10. modphysnoob
    • one year ago
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    ok

  11. modphysnoob
    • one year ago
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    so left diagonal would be {1,1}, not sure if I need to normalize it? {xy,-x^2-y^2} doing dot product would result in same in scalar {xy -x^2-y^2} Integreate[xy-x^2-y^2 ] x=0 to 1 , y =0 to 1

  12. JamesJ
    • one year ago
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    Do the vertical piece next. The easiest way to do this and avoid confusion is to explicitly parameterize the curve over which you are integrating. For the vertical part, you are integrating over the portion of the line (1,t), where t = 0 -> 1 Hence the integral over that line is \[ \int_0^1 -(x^2 + 2y^2) \ dt = \int_0^1 -(1 + 2t^2) \ dt \] For the diagonal, use the same idea of parameterization.

  13. modphysnoob
    • one year ago
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    oh, I am doing 3.51, which uses the figure 3.50)b)

  14. JamesJ
    • one year ago
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    Ok, in which case a perfectly good parameterization for the line from (2,0) to (1,1) is (2-t,t), t = 0 -> 1

  15. modphysnoob
    • one year ago
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    so in that case , you are assigning 2-t to x and t to y right? but what happen to dot product between E and dl

  16. JamesJ
    • one year ago
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    dl = (-i + j) dt

  17. JamesJ
    • one year ago
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    Sounds to me like you should go back to your maths textbook or something and review line integrals

  18. modphysnoob
    • one year ago
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    I took calc 3 long time ago, never thought it would be useful in real life

  19. JamesJ
    • one year ago
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    I'll help finish this part and then I'm out of here. E(t) = xy i + -(x^2 + 2y^2) j = (2-t)t i - ( (2-t)^2 + 2t^2 ) j Now calculate E(t). dl and integrate that expression from t = 0 to 1

  20. modphysnoob
    • one year ago
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    Integrate[(2-t)t+(2-t)^2+2t^2,{t,0,1}

  21. JamesJ
    • one year ago
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    No, dl = (-i + j) dt hence E(t) . dl = -(2-t)t + - ( (2-t)^2 + 2t^2 )

  22. modphysnoob
    • one year ago
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    ok, thanks

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