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Electric force acting on a single charge

  1. Nov 19, 2009 #1
    1. The problem statement, all variables and given/known data

    http://img682.imageshack.us/img682/985/physicscharge.png [Broken]

    2. Relevant equations
    3. The attempt at a solution

    http://img682.imageshack.us/img682/4295/eqn.png [Broken]

    Is it the first one or the 2nd one? Is my basic concept right in the first place?
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Nov 20, 2009 #2

    Redbelly98

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    Your first equation is the correct one. Q3 repels Q1, and thus exerts a negative (to the left) force on Q1.
     
  4. Nov 20, 2009 #3
    I disagree, unless you're talking about the absolute value of the charges. The way the first one is written, all 3 forces will be pointing towards the negative x direction.

    Set the force vectors all in the same direction, and let the sign of the charge resolve whether it attracts or repels.

    In your example, each force vector should point from Q2, Q3 and Q4 to Q1. If the signs are the same, it's a repulsive force, and the vector stays pointing the same way. If the signs are opposite, it is a negative number, and that force points the other way.

    But initially, have all vectors pointing in the same direction. If you're measuring the force caused by 2 on 1, point the vector from 2 to 1. If it turns out to be attractive, the negative sign will take care of it.
     
  5. Nov 20, 2009 #4

    Redbelly98

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    True, I was thinking in terms of the absolute values of the charges.

    But the best way to use Coulomb's law is with the absolute values of the charges, then use the "opposites attract, likes repel" rule to figure out the direction.

    The first equation, with absolute value signs on the charges, would be correct. The second equation is still wrong, whether it's absolute values or actual +/- charge values.
     
  6. Nov 20, 2009 #5
    I don't know, that still sounds like a sloppy way to do it. What happens when the charge is an unknown test charge?

    Setting up a force analysis, and just picking a direction has always worked for me. If the answer comes out negative, I picked the wrong direction.

    The second equation used will give the correct magnitude of the force. Multiplying each force by the -x unit vector will give the correct answer.

    But you're right, the first equation with absolute value signs will provide the correct answer. I personally just think it's easier to write the general algebraic expression for the forces first in generic terms, then fill in numbers (including negative signs) later.
     
  7. Nov 20, 2009 #6

    Redbelly98

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    Well, it gets tricky if you are doing the force on one of the middle charges, say Q2 = -2.0C. In this problem, Q1 and Q3 are both positive. Blindly using using Coulomb's law would say the forces from Q1 and Q3, on Q2, are both negative, hence in the same direction. In fact Q1 attracts Q2 to the left, while Q3 attracts Q2 in the opposite direction, to the right.

    For that reason, I prefer to think about the force direction for each charge pair individually.

    I always assume a positive test charge. If I need to plug in a negative charge value later on, I know to reverse the direction of the force.
     
  8. Nov 20, 2009 #7
    thanks guys.

    so from above; i assume it will be correct if i absolute the charges in the first equation and for the 2nd eqn to remain unchanged?

    I got a positive ans for the first eqn and a -ve for the 2nd eqn. so what does it means?


    I dun quite understand this part, base on the assumption u mentioned, how do i apply it into the eqn?
     
    Last edited: Nov 20, 2009
  9. Nov 20, 2009 #8
    Just follow Redbelly's advice. It's probably easier for most people to understand.
     
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