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Force-neutralizing charge (electrostatics)

  1. Jun 12, 2016 #1
    1. The problem statement, all variables and given/known data
    A charge 2Q is placed a distance 'd' from charge 4Q. A third charge 'q' is placed 3m from charge 2Q directly on the line between charges 2Q and 4Q. Find d and q such that the force between charges 2Q and 4Q is equal to 0.
    2. Relevant equations

    Coulomb's law: f = k(|q1q2|)/r2

    k = 8.99 x 10^9 (Nm^2/c^2)

    3. The attempt at a solution
    whatisthis_zps2z2kj2tw.png

    (k)(2Q)(q)/3 = [k(4Q)(q)]/(d - 3)

    (d - 3)/3 = 2

    d - 3 = 6;

    d = 9m?


    k(2Q)(q)/3 + [k(4Q)(q)]/6 = k(2Q)(4Q)/9;

    q[(k4Q)/6 + (k4Q)/6] = k8Q2/9;

    q[k4Q/3] = k8Q2/9;

    q = (3)(2Q)/9 ?

    q = 2Q/3 ?
     
  2. jcsd
  3. Jun 12, 2016 #2

    cnh1995

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  4. Jun 13, 2016 #3
    Thanks for catching that.

    (d - 3)2/3 = 2;

    (d2 - 6d + 9)/3 = 2

    d2 - 6d + 9 = 6

    d2 - 6d + 3 = 0

    Quadratic formula is:
    -b +- (b2 - 4a)1/2/2a

    (6 +- (24))1/2/2

    d = 30 or d = -18?
     
  5. Jun 13, 2016 #4

    cnh1995

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    There should be 9 in the denominator instead of 3.
     
  6. Jun 15, 2016 #5
    Thanks, this isn't going so well for me :) :

    (d - 3)2/9 = 2;

    (d2 - 6d + 9)/9 = 2

    d2 - 6d + 9 = 18

    d2 - 6d - 9 = 0

    I actually made a mistake for the quadratic formula part too in my last post:
    -b +- (b2 - 4a)1/2 / 2a

    d = [6 +- (24 - 4)1/2]/2

    d = [6 +- (4.472)]/2

    d = 5.23 or d = 0.764 ?
     
  7. Jun 15, 2016 #6

    cnh1995

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    It's (b2-4ac)1/2.
    You'll get only one positive answer. Be careful with the signs and square terms.
     
  8. Jun 16, 2016 #7
    Still not going well for me then..

    -b +- (b2 - 4ac)1/2 / 2a

    d = [6 +- (24 - (4)(1)(-9))1/2]/2

    d = 6 +- (24 + 36)1/2 / 2

    d = 6 +- (60)1/2/2

    d = (6+- 7.745)/2

    d = 6.8725 or d = -0.8725 ?

    I'm going to assume use only the positive answer for d ? :

    [k(2Q)(q)]/9 + [k(4Q)(q)]/(6.8725 - 3)2 = [k(2Q)(4Q)]/81;

    [k(2Q)(q)]/9 + [k(4Q)(q)]/15= [k(2Q)(4Q)]/81;

    q[(k4Q)/15 + (k2Q)/9] = (k8Q2)/81

    q[(3)(k)(4Q) + (5)(k)(2Q)]/45 = (k8Q2)/81

    q[(12)(k)(Q) + (10)(k)(Q)]/45 = (k8Q2)/81

    q[k22Q/45] = (k8Q2)/81

    q = [(45)(k8Q2)]/[(81)(k22Q)];

    q = [360kQ2 /(1782kQ) ;

    q = 0.202Q ?
     
  9. Jun 16, 2016 #8

    cnh1995

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    b2=36 and not 24..
     
  10. Jun 19, 2016 #9
    I forgot the '-b' at the beginning as well...

    Hopefully everything is more neat now:

    -b +- (b2 - 4ac)1/2 / 2a

    d = [-6 +- (62 - (4)(1)(-9))]/(2)

    d = [-6 +- (36 + 36)]/2

    d = [-6 +- (72)]/2

    d = 33 or d = -34?

    Assuming using only the positive answer again:

    [k(2Q)(q)]/32 + [k(4Q)(q)]/(33 - 3)2 = [k(2Q)(4Q)]/81;

    [k(2Q)(q)]/9 + [k(4Q)(q)]/900 = [(k)(2Q)(4Q)]/81;

    [(100)(k)(2Q)(q)]/900 + [(k)(4Q)(q)]/900 = [(k)(2Q)(4Q)]/81;

    [200Qkq + 4Qkq]/900 = [(k)(2Q)(4Q)]/81;

    204Qkq/900 = [(k)(2Q)(4Q)]/81; now let's simplify the right hand side:

    204Qkq/900 = [(k)(8Q2)]/81

    q = [(900)(8Q2)(k)]/[(204)(81)(Q)(k)]

    q = (7200Q)/(16524)

    q = 0.435Q
     
  11. Jun 19, 2016 #10

    cnh1995

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    How come the test charge is only 3m away from 2Q but 30m away from 4Q?
    In the end, put your value of d in this expression and see if it satisfies the equation.
     
    Last edited: Jun 19, 2016
  12. Jun 21, 2016 #11
    -b +- (b2 - 4ac)1/2 / 2a

    d = [-6 +- (62 - (4)(1)(-9))1/2]/(2)

    d = [-6 +- (36 + 36)1/2]/2

    d = [-6 +- 8.4852]/2

    d = 2.485 or d = -14.48 ?


    (k)(2Q)(q)/32 + (k)(4Q)(q)/(2.485 - 3)2 = (k)(2Q)(4Q)/(2.485 + 3)2;

    (k)(2Q)(q)/9 + (k)(4Q)(q)/0.265225 = (k)(8Q2)/30;

    q[(137.72)(k)(Q)]/9 = (k)(8Q2)/30;

    q = [(k)(9)(8Q2)]/[(30)(137.72)(k)(Q);

    q = 0.01742Q
     
    Last edited: Jun 21, 2016
  13. Jun 21, 2016 #12

    cnh1995

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    -b=6 and not -6. It will give you
    d=(6+8.4852)/2=7.2426 m.
    You can verify the answer by putting it in this equation.
     
  14. Jun 29, 2016 #13
    Apologies for bumping this late I will try to be more active now. So in the beginning I had (corrected as) this:

    2Q*q*k/32 = 4Q*q*k/(d - 3)2;

    (d - 3)2/9 = 2;

    And then I ended up later with this polynomial:

    d2 - 6d - 9 = 0;

    Then used the quadratic formula:

    -B +- (B2 - 4ac)1/2/2a;

    where a = 1, b = -6, and c = -9

    d = 6 +- (36 + 36)1/2/2;

    d = 6 +- (72)2/2;

    d = 6 +- (8.4852)/2;

    d = 7.2426 or d = -1.2426; and verifying it: (7.2426 - 3)2/9 = 2 -> 17.999/9 = almost 2, which would check out.


    So what do I do about q now again? I get this doing it over again:

    d = 7.2426;

    (k)(2Q)(q)/32 + (k)(4Q)(q)/(7.2426 - 3)2 = (k)(2Q)(4Q)/(7.2426)2 which i corrected as it is d that was just the entire distance across;

    Now factoring out q I get this:

    q[k(2Q)/9 + k(4Q)/18] = k(8Q2)/52.4552 ;

    q[k(4Q)/18 + k(4Q)/18] = k(8Q2)/52.4552 ;

    q[k(8Q)/18] = k(8Q2)/52.4552 ;

    q[k(8Q)/18] = k(8Q2)/52.4552;

    q = [18 / 52.4552]*[k(8Q2)/k(8Q)] ;

    q = [18 / 52.4552]Q;

    q = 0.3431Q ?
     
  15. Jun 29, 2016 #14

    haruspex

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    I'm not sure I understand the question, and I certainly do not understand the attempted solution.
    What is meant by "the force between charges 2Q and 4Q"? The obvious meaning is the force they exert on each other, but that would be unaffected by q. You seem to have interpreted it as the net force exerted on q, but it cannot be that since you would have no way to determine q.
    How do you arrive at the equation here:
    You have the two forces exerted by 2Q and 4Q on q adding instead of opposing, and why should they add to the force 2Q and 4Q exert on each other?
    A possible interpretation is that there should be no net force on either 2Q or 4Q. That would give you two equations with two unknowns. But if that is what the question means the wording is very poor.
     
  16. Jul 2, 2016 #15
    Yes, this.

    I made it that way because it was across the whole distance, but I probably should have made '(k)(2Q)(4Q)/(7.2426)2' negative, would this make more sense?

    (k)(2Q)(q)/32 + (k)(4Q)(q)/(7.2426 - 3)2= -(k)(2Q)(4Q)/(7.2426)2
     
  17. Jul 2, 2016 #16

    haruspex

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    Your equations still do not make any sense. Start again from scratch: what are the forces acting on the 2Q?
    (You can drop the factors of k; clearly they will always cancel out.)
     
  18. Jul 4, 2016 #17
    Okay, I think I know what you mean now. I really might need a refresher on basic mechanics (and everything, really):

    Do I sum up the forces then, on each charge, like so?:

    k(2Q)(q)/32 + k(4Q)(2Q)/d2 = k(2Q)(4Q)/d2 + k(q)(4Q)/(d - 3)2
     
  19. Jul 4, 2016 #18

    haruspex

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    That's better, but what you have written is that the net force on the 2Q is equal to the net force on the 4Q. According to the interpretation you confirmed in your post #15, we know a bit more than that.
     
  20. Jul 4, 2016 #19
    Okay so the forces on each charge are in opposite directions and on one side will be negative then, like this?:

    k(2Q)(q)/32 + k(4Q)(2Q)/d2 = -[ k(2Q)(4Q)/d2 + k(q)(4Q)/(d - 3)2 ]
     
  21. Jul 4, 2016 #20

    haruspex

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    No, you misunderstand.
    As you agreed, the condition is that the net force on each of 2Q and 4Q (separately) is zero.
    So just consider the net force on one at a time. Don't try to combine them into one equation.
     
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