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Find the point b/w two charges where the electric field is zero?

  1. Aug 25, 2013 #1
    1. The problem statement, all variables and given/known data

    Two positive charges of 16 micro coulomb and 4 micro coulomb are separated by a distance 3 m along a straight line....find the spot on the line joining the two charges where Electric field is zero?

    2. Relevant equations

    E = F/Q and F = Q1Q2/R^2

    3. The attempt at a solution

    AS i tried it i thought electric field is zero when forces b/w two charges will become equal,i took a point P on a line,so that E2 is a distance "d" from P,and E1 is at a distance "3-d" from p...so i applied F1=F2 but the answer came out to be 3 which wasn't actually,i checked the textbook and they equalled E1=E2


    actually i know its solution but couldn't grasp the idea of the difference b/w electric field/coloumb's law and electric force? i couldn't help thinking my self it can also to be F1 = F2 but why it is E1 = E2??? what is the idea notion behind it? please can anyone explain it in layman's terms?
     
  2. jcsd
  3. Aug 25, 2013 #2
    Force between two charges is ALWAYS equal and opposite by Newton's 3rd law.
    Thats why you are getting the same distance as given in the question.
    Electric field is given by the force on a small positive test charge. The force on this charge (or any other charge for that matter) is zero when electric field is zero.
     
  4. Aug 25, 2013 #3
    can you elaborate it a bit more please sir
     
  5. Aug 25, 2013 #4

    CAF123

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    You do not want E1=E2 here, you want E1+E2=0.
     
  6. Aug 25, 2013 #5
    how?
     
  7. Aug 25, 2013 #6

    haruspex

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    You don't care about the force of the two charges on each other. The field relates to the force that would be experienced by some other charge placed at that point. So if you want to do it by balancing the forces, imagine introducing a 'test' charge q at distance d from one and 3-d from the other and calculate the two forces exerted on it.
     
  8. Aug 25, 2013 #7

    CAF123

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    The E field from a positive charge is radially outward and proportional to 1/r2. So only for large enough r does E tend to zero. Solving like E1=E2, is saying I want the E field from charge 1 to be zero and I want the E field from charge 2 to be zero and then solving for x. But there is no x.

    If you draw a picture and sketch the E field lines, you can see there is a point between the two charges were the total E field is zero, total meaning the superposition of the E fields from both sources.
     
  9. Aug 25, 2013 #8
    couldn't get the answer by F1= -F2 or F1 + F2 = 0..please help
     
  10. Aug 25, 2013 #9

    CAF123

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    The equation is correct, but would you be able to show more of your work?
     
  11. Aug 25, 2013 #10
    (Kq1q2)/(d^2) + (kq1q2)/(3-d)^2 = 0

    kq1q2 can be taken as a common factor,so it will be eliminated...

    1/d^2 + 1/(3-d)^2 = 0

    Taken L.C.M,as RHS = 0 Denominator will become 1,

    (3-d)^2 + d^2 = 0

    9 + d^2 - 6d + d^2 = 0

    2d^2 - 6d + 9 = 0

    now how to factor that?
     
  12. Aug 25, 2013 #11
    also sir please i am still not able to grasp the notion of making E1 = E2,couldn't comprehend the question clearly :( please elaborate in the simple possible terms if you can sir
     
  13. Aug 25, 2013 #12

    CAF123

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    What is q1 and what is q2? You want to find the point between the 16μC and the 4μC where the E field is zero. So you do not need to consider forces at all. You can do it with forces however. A test charge placed at the location of zero E field would experience zero net force. There are two forces acting on the test charge: the force from the 16μC charge and the force from the 4μC. We do not need to be concerned with the force between these two charges.

    Put the 16μC at the origin for simplicity. The test charge is a distance d from the 16μC and a distance 3-d from the 4μC charge.
    What is the force acting on the test charge due only to the 16μC?
    What is the force acting on the test charge due only to the 4μC?

    Note their directions too. Like repels like and take the test charge to be positive. (it could equally have been negative).
     
  14. Aug 25, 2013 #13

    haruspex

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    What are q1 and q2? With the forces approach, you have three charges: the two given charges (call those q1 and q2) and a test charge q. You want the force between q1 and q balancing the force between q2 and q.
     
  15. Aug 26, 2013 #14
    couldn't get that :(
     
  16. Aug 26, 2013 #15

    haruspex

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    An electric field exerts a force on a charge placed within the field. Where the field is zero, the force is zero.
    You are asked for a locus at which the net field from two given charges is zero. That means that a 'test' charge (i.e. some third charge) placed at that point will feel no net force. That is, the two forces it feels from the two given charges balance out.
    If the two given charges are q1 and q2, placed distance s apart, and a third charge q is placed distance d from q1 and distance s-d from q2:
    - what force is exerted on it by q1?
    - what force is exerted on it by q2?
    - what is the magnitude of the net force?
     
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