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JenniferSmart1

  • one year ago

Question: I solved the first part and I just need help understanding the second part. This will take me a a few min to write and draw....bear with me...

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  1. JenniferSmart1
    • one year ago
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    Point charges of \[q_1=-5.00 \mu C\]\[q_2=+3.00 \mu C\]\[q_3=+5.00\mu C\] located on the x access at \[x=-1.00cm\]\[x=0\]\[x=+1.00cm\]respectively. I calculated \[\vec{E}\] on the x axis atx=3.0cm and x=15.0cm and got: \[\vec{E}_{p1}=(1.14\times10^8N/C)\hat{i}\] and \[\vec{E}_{p1}=(1.74\times10^6N/C)\hat{i}\] Here is where I'm having trouble: ARe there any points on the x-axis where the magnitude of the electric field is zero? If so, where are those points?

  2. JenniferSmart1
    • one year ago
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    @Luis_Rivera @phi

  3. JenniferSmart1
    • one year ago
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    typo: I meant to write \[\vec{E}_{p2}=(1.74\times10^6N/C)\hat{i}\]

  4. JenniferSmart1
    • one year ago
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    @Luis_Rivera I'll post the answer, can you help me understand it? I have the solution but I'm having a hard time understanding the concept

  5. JenniferSmart1
    • one year ago
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    @Jemurray3

  6. Jemurray3
    • one year ago
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    |dw:1360443203643:dw|

  7. Jemurray3
    • one year ago
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    Think of it this way: if you were just slightly to the left of the particle at x = -1, the electric field would point to the right, correct?

  8. JenniferSmart1
    • one year ago
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    correct

  9. Jemurray3
    • one year ago
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    Now go a hundred billion miles to the left. The distances between the three would be almost zero, so you'd really only be able to see a single particle of charge -5 + 3 + 5 = 3. Does that argument make sense?

  10. JenniferSmart1
    • one year ago
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    Yes, so we would see a particle of charge three. Oh so is \[r\rightarrow \infty\]?

  11. Jemurray3
    • one year ago
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    No. The point of that is if you're close to the one at -1, then the field points to the right. If you're far away, the field points to the left, so there must be some point at which the field is zero.

  12. JenniferSmart1
    • one year ago
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    I guess that kinda makes sense....where from here?

  13. Jemurray3
    • one year ago
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    The fields from the three particles are: \[E_1 = \frac{-5}{(x+1)^2} \] \[E_2 = \frac{3}{x^2} \] \[E_3 = \frac{5}{(x-1)^2} \] So the equation you should be solving is \[E_{total} = E_1 + E_2 + E_3 = 0 \]

  14. JenniferSmart1
    • one year ago
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    \[E_1=\frac{q_1}{(x+1)^2}\] this means: the electric field due to a point charge from a positive test particle "x" located at a location (which I'm solving for) \[E_{total} = E_1 + E_2 + E_3 = 0\] by solving for x, I've found a place on the x axis where the electric field due to all point charges is zero....

  15. JenniferSmart1
    • one year ago
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    makes sense. THanks @Jemurray3 !

  16. JenniferSmart1
    • one year ago
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    The positive and negative charges are bothering me...|dw:1360445236763:dw|

  17. Jemurray3
    • one year ago
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    What's bothering you?

  18. JenniferSmart1
    • one year ago
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    I Would've written this as: \[-\frac{q_1}{(x-1)^2}-\frac{q_2}{x^2}-\frac{q_3}{(x+1)^2}=0\]

  19. JenniferSmart1
    • one year ago
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    because the charge on q_1 is negative

  20. Jemurray3
    • one year ago
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    You must also take into account the direction the field is pointing. The field from the q_1 charge is to the right.

  21. Jemurray3
    • one year ago
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    Generally speaking \[ \vec{F} = \frac{kQ}{r^2} \hat{r} \] So Q is negative, but that is compensated by the direction of r hat.

  22. JenniferSmart1
    • one year ago
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    That explains the positive sign, what about the x_1 part? why do they have it as x-(-1)?

  23. JenniferSmart1
    • one year ago
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    I mean the x-1 part

  24. Jemurray3
    • one year ago
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    That is the distance between a point (x) and the point (-1).

  25. JenniferSmart1
    • one year ago
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    which should be x-1 though. That doesn't make sense in practice...I think.... Ok let's see (I can't believe I'm struggling with simple arithmetic) \[\triangle x=|x_2-x_1|\] so let's say our point is at x_2=-5 Oh I see....nevermind haha yep it's plus (+)

  26. Jemurray3
    • one year ago
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    If you ever forget, just say "okay, so where is the distance zero?" and if it matches the location of the actual particle then you're good.

  27. Jemurray3
    • one year ago
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    so (x+1) is zero at x=-1 , which is where the particle is.

  28. JenniferSmart1
    • one year ago
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    yep, I'll remember that. Thanks again @Jemurray3 !!!

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