I am unhappy about the answer to this problem

  1. 1. The problem statement, all variables and given/known data

    [​IMG]

    In the figure above, determine the point (other than infinity) at which the electric field is zero. Given q1 = -2.50[tex]\mu[/tex]C and q2 = 6.00[tex]\mu[/tex]C

    Solution: 1.82m to the left of q1


    3. The attempt at a solution

    So I set it up as

    [tex]\frac{q_{1}}{x^2} = \frac{q_{2}}{(d-x)^2}[/tex] where is x is the point I am looking for

    After some simplification

    [tex]q_{1}d^2 - 2dq_{1}x + x^2(q_{1} - q{2}) = 0 [/tex]

    Here is what hit me, if I was only going to concern about the magnitude, I would get x = -1.82m and 0.39m

    First of all, I am not even sure why it is -1.82m and not 0.39. Isn't -1.82m outside the range of d = 1.00m?

    Secondly, if I do not concern with the magnitude (using -2.50µC instead of 2.50µC), I couldn't even solve the quadratic.

    So why must I use positive 2.50µC? And why is it -1.82m and not 0.39m?

    Thanks!
     
  2. jcsd
  3. Pengwuino

    Pengwuino 7,118
    Gold Member

    Recall that an electric field will give rise to a force on this charged particle you plan on introducing at some position 'x'. So you've sent to find where the field and thus, the force would vanish. If it is indeed 0.39m to the right of q1... how would that physically be possible?

    If you introduced a test charge that is negative, it would receive a pull from q2 and a push from q1 - obviously not a point of vanishing electric field. If the test charge is positive, the opposite occurs - the charge receives a push from q2 and a pull from q1. This point can't be a point with no electric field.

    Now consider the -1.83m position (or really, just look at the left side of q1) and run through the same arguments and hopefully you can see why the charge has to be on the left side of q1.

    The charge does not have to be placed in between the charges. Infact, if q1 and q2 are of opposite charge, there exists no point between the charges with 0 electric field by the argument above. If q1 and q2 are of the same charge, the only two points of 0 electric field are in between q1 and q2.
     
  4. Of course it could!

    [​IMG]

    If it is to left for real

    1) A positive test charge would be attracted to the negative charge q1, but the charge q2 is greater, but also further and hence would make a "0" net charge

    2) A negative test charge would be repelled by q1, at the same time attracted to q2
    But, just going back to my quadratic, why does only magnitudes of the charge work?
     
  5. gneill

    Staff: Mentor

    But q2 is a positive charge. It would repel your test charge, pushing it towards q1 which is also attracting it. It would not be a "0" net field (not charge!). It would not be a point of no force for the positive test charge.

    Right. So not a zero force point for the negative test charge either.

    You're already encapsulating the geometry of the situation into the equation. In particular, you're taking into account the directions of the fields caused by each charge in relation to their placement when you write the equation the way you did.

    If you wanted to make your life more difficult you could write an equation that would make no assumptions about polarity of charge and the resulting force directions as you move from one side of a charge to another. This would necessarily be a vector equation. Then you could spend a quiet hour or two figuring out how to work vectors into a quadratic equation...
     
  6. Could you dumb that last bit for me?
     
  7. remember the electric field of a positive charge flows outwards and negative flows inwards. so you should have to be on the right of 2 or on the left of 1 to cancel out the e field


    [​IMG]

    q1 and q2 are of different magnitudes and different sign
     
  8. Pengwuino

    Pengwuino 7,118
    Gold Member

    Don't confuse force with charge.

    If the charged particle is placed between Q1 and Q2 being pushed from Q1 and pulled from Q2 (thus, a negative test charge), the forces are adding and the result is a particle accelerating to the right. With a positive test charge, it would be pulled to the left and pushed from the right.
     
  9. my bad, its 2am... ignore my post then
     
  10. Delphi51

    Delphi51 3,410
    Homework Helper

    In your first equation in the first post, the (d - x) assumes the point is between the charges. Use (d + x) to get the answer.
     
  11. SammyS

    SammyS 8,884
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    Hi flying pig.

    To the left of q1, which is at x = 0, the electric field due to q1 is:

    [tex]E_1=-k\frac{q_1}{x^2}\,.[/tex]

    To the right of q1, the electric field due to q1 is:

    [tex]E_1=k\frac{q_1}{x^2}\,.[/tex]


    Similarly, to the left of q2, which is at x = d, the electric field due to q2 is:

    [tex]E_2=-k\frac{q_2}{\left(x-d\right)^2}\,.[/tex]

    To the right of q2, the electric field due to q2 is:

    [tex]E_2=k\frac{q_2}{\left(x-d\right)^2}\,.[/tex]

    The vector nature of the electric field in the above, is indicated by the signs. The effect of the signs of the charges on the electric field is built in.

    The problem asks you to find a finite value of x at which the electric field is zero. That is:

    [tex]\text{Solve }E_1+E_2=0[/tex]

    To the left of both charges and to the right of both charges:

    [tex]k\frac{q_1}{x^2}+k\frac{q_2}{\left(x-d\right)^2}=0\quad\to\quad \frac{q_1}{x^2}=-\frac{q_2}{\left(x-d\right)^2}\,.[/tex]
    The RHS has the opposite sign of the RHS of your equation. However, IF you only consider magnitude (Why would you do that?), this is equivalent to your equation. Of course, the only way for this equation to be true, is for q1×q2<0.

    Between the charges:

    [tex]k\frac{q_1}{x^2}-k\frac{q_2}{\left(x-d\right)^2}=0\,.[/tex]
    This only has a solution if q1×q2>0.
     
  12. It's more confusing if you use x - d even though the end result is the same because of the square

    But why would I use (d+x)? I am really confused.
     
  13. Delphi51

    Delphi51 3,410
    Homework Helper

    Sorry! I was thinking of x as a positive number. Your d-x is great; x just comes out to a negative number to indicate a point left of the origin.
     
  14. I read Sammy's post over and over again which seems to be answering my question, but it just doesn't hit my head yet.
     
  15. Pengwuino

    Pengwuino 7,118
    Gold Member

    The math doesn't care about the physics. One of the skills you must build up is determining what solutions make sense and which do not. This is kind one of those cases.

    Also, I didn't notice it but you asked why you must use the magnitudes of the charges. Well, for one, you equate two equations with strictly positive denominators. If one side has a charge that is negative and one that is positive, there's no way they can be equal simply because one is always positive and one is always negative.

    Compute the forces at the point near the middle. They will be equal but in the same direction so it's not a solution.
     
  16. Sammy made me just realize something last night.

    The forces are attractive and hence x can and should never be x = 0.39m even though the math worked out because I am saying there exists a charge in the middle such that the E-field is 0. But what does x = 0.39m mean then?

    So when the charges are repulsive, it would be right to use d - x, but otherwise, d + x?

    That makes so much sense I didn't even see it, I will now repeat the fourth grade!
     
  17. Delphi51

    Delphi51 3,410
    Homework Helper

    I played with this some more and am surprised at how complicated it is!
    I started with q1/x² + q2/(d-x)² = 0 and it worked perfectly, using the q1 = -2.5 and q2 = 6 and producing x = -1.82. Also a positive x spurious answer.

    BUT if q1 happened to be a positive number so the solution is an x between 0 and 1, it would NOT work because both the q1/x² and the q2/(d-x)² are positive as if the E field from both charges is to the right. There should be a solution between 0 and 1 but my equation cannot find it.

    I cannot think of a way to write the E1 + E2 = 0 equation that works for solutions in all three regions of x. The physicist must determine the zone where the solution lies before writing the equation!
     
  18. Alright, I set it up again, but I still can't solve it

    [tex]-\frac{|q_{1}|}{x^2} = \frac{|q_{2}|}{(d+x)^2}[/tex]

    [tex]-|q_{1}| + 2x|q_{1}|+ x^2(|q_{1}|- |q_{2}|) = 0[/tex]
     
  19. How did you use the equation and used negative signs of the charges? It only works when you use the magnitude and I think that is why I got it wrong. Because when I set them equal to each other, that already took care of the signs.

    But why did you use (d-x)? It really should be d +x


     
  20. Delphi51

    Delphi51 3,410
    Homework Helper

    My slightly different equation worked perfectly with the -2.5 and +6.
    I was thinking E1 + E2 = 0 and E1 was negative while E2 was positive, so it worked out.

    But it wouldn't work in a situation where both charges are positive. In that case there should be a solution with x between 0 and 1, but my equation fails - both terms are positive and can't add up to zero.

    These equations do not take care of the signs for the E1 and E2 terms properly, due to the fact that squaring the distance wipes out its sign.
     
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