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Net positve charge

  1. May 2, 2017 #1
    1. The problem statement, all variables and given/known data
    Consider a high voltage direct power line which consist of two conductor suspended 3 meters apart. The lines are oppositely charged, If electric field halfway between them is ##15000N/C## how much is the excess positive charge on a 1 km length of positive conductor ?

    2. Relevant equations

    ##\displaystyle E = {2Q \over y \sqrt{4y^2 + l^2}}##

    where ##l## is length of wire, ##y## is the distance of the charge from the wire and the wire is kept on the x-axis.

    3. The attempt at a solution

    Electric field between the conductors will be sum of indiviual E fields, so I calculated indiviual E field and then took the sum.

    ##\displaystyle E_{+ve} = {4kQ \over 3 \sqrt{9 + 10^6}}##

    ##\displaystyle E_{-ve} {-4kq \over 3\sqrt{9 + 10^6}}##

    Solve for ##Q - q## in ##E = E_{+ve} + E_{-ve}## I get ##1.25 \times 10^{-3} C##

    But the given answer is ##6.26 \times 10^{-4} C##

    Am I missing something ?
     
  2. jcsd
  3. May 2, 2017 #2

    TSny

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    You are considering E at a point halfway between the two lines. How far is this point from each of the lines?

    This distance is very small compared to the length of the lines. So, it is safe to treat the lines as "infinitely long".

    Also, I believe the two lines are considered to have equal but opposite charges. Otherwise, the problem would not have a definite answer.
     
  4. May 3, 2017 #3
    it seems like two infinity line charge,,,
    at an infinity line charge, E is ##E = \frac {2 k λ} {a} ##
    where λ is charge per length, and a is distance perpendicular from the line,
    there're two oppositely lines, so ##2 E = 15000N/C## ,

    Just for suggestion, by subtituting a, then from magnitude of λ, it would be determined the charge that excessed,,,
     
  5. May 3, 2017 #4
    It does not matter much, I tried that also. I get roughly same as above.

    I used ##E_{+ve} = {2k\lambda \over R} = 12\times 10^6 Q##

    and ##E_{-ve} = -12 \times 10^6 q##.

    I get the exact same answer when I equate ##E_{+ve} + E_{-ve} = 15000##
     
    Last edited: May 3, 2017
  6. May 3, 2017 #5

    TSny

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    Are you now getting the correct answer? If not, are you using the correct distance for R?
     
  7. May 3, 2017 #6
    No I don't. I am using ##3/2## for the distance, I think it is correct.
    Should I post complete calculation ?
     
  8. May 3, 2017 #7

    TSny

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    OK
    Yes, please.
     
  9. May 4, 2017 #8
    ##\lambda = \dfrac{Q}{1000}##

    ##E_{+ve} = \dfrac{2k\lambda}{R} = \dfrac{2 \times 9 \times 10^9 }{3/2} \cdot \dfrac{Q}{1000} =## 12 x 10^6 Q

    Similarly, we replace ##-q## for ##Q## and get ##E_{-ve} = -12 \times 10^6 q##

    ##E_t = E_{+ve} + E_{-ve} =12 \times 10^6 Q - 12 \times 10^6q \implies 1.5 \times 10^4= 12 \times 10^6 (Q - q) \implies Q -q = 1.25 \times 10^{-3} ##
     
  10. May 4, 2017 #9

    TSny

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    You can assume that the charge on the negative line is equal in magnitude to the charge on the positive line. So, If Q is the charge on 1000 m of the positive line, then the charge on 1000 m of the negative line is -Q.

    At a point halfway between the two lines, how does the magnitude of the electric field from the negative line compare to the magnitude of the electric field from the positive line?

    How does the direction of the electric field from the negative line compare to the direction of the electric field from the positive line?
     
  11. May 4, 2017 #10
    If I am assuming ##Q =q## then how do we get excess positive charge ?

    Won't then ##E_{+ve} = 12 \times 10^6 Q## and ##E_{-ve} = -12 \times 10^6 Q##

    So ##E_t = 0 ## no ?

    I am a bit confused now.
     
  12. May 4, 2017 #11

    TSny

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    One of the power lines has a net positive charge. This net positive charge is the "excess positive charge" on that line. Likewise, the other line has "excess negative charge". The total charge of both lines together is zero. The question is asking for the excess charge on 1000 m of just the positively charged line.

    Here you want to make sure to sketch a picture showing the parallel power lines and pick a point P midway between the lines. At P, draw a vector representing the electric field produced by the positively charged line. Then draw another vector at P for the electric field produced by the negatively charge line. Do the two electric field vectors at P point in the same direction or the opposite direction?
     
  13. May 4, 2017 #12

    ##E_{+ve} = 12\times 10^6Q (-\hat j)##
    ##E_{-ve} = -12\times 10^6Q (-\hat j)##
    ##E_{t} = 0##

    afasfasc.png

    No ?
     
  14. May 4, 2017 #13

    TSny

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    When you want to represent an electric field vector at a point P, the vector should be drawn with the tail of the vector at P. So, the vector representing the electric field at P due to the positively charged line should be drawn with the tail at P. Likewise for the electric field at P due to the negative power line. So, you should draw both the blue and red arrows starting on point P.

    upload_2017-5-4_10-24-57.png
     
  15. May 4, 2017 #14
    I would still draw both red and blue arrow going downwards (in same direction).

    I think I should draw them in opposite direction to get the answer but I don't know why I should do that.
     
  16. May 4, 2017 #15

    TSny

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    You are right that both arrows would point downward in the same direction. When you add those two vectors together, do the two vectors cancel to give a total field of zero?
     
  17. May 4, 2017 #16
    ahh, I should get a vector of double length.

    ##E_{t} = (-\hat j) Q \times 10^6 (12 + 12) = 24 \times 10^6 Q (-\hat j) = 15 \times 10^3 (-\hat j ) \implies 6.25 \times 10^{-4}##

    Are we done ?
     
  18. May 4, 2017 #17

    TSny

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    Yes, that looks good!
     
  19. May 4, 2017 #18
    Gosh thank god !!
     
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