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Conditions for neutral point to exist

  1. Oct 9, 2015 #1
    I want to verify whether I am correct .As the title suggests It is about point where potential is zero (point other than infinity)
    Note:x is distance of neutral point from charges
    Firstly,I would like to ask
    Is coordinate system necessarily be there
    1) If there are two positive charges for the neutral point to exist one of them should have negative x,so that one of the two potentials becomes negative and the two potentials should have same magnitude so that they cancel each other.Right?
    And in order to have negative x ,coordinate system must be present.
    2)If there are two negative charges for the neutral point to exist one of them should have negative x,so that one of the two potentials becomes positive(negative sign of q and r would cancel each other) and the two potentials should have same magnitude so that they cancel each other.Right?
    3)If there are two charges,one positive and the other negative ,then in this case for the neutral point to exist both of them (charges)should have positive x ,and the two potentials should have same magnitude so that they cancel each other.Right?
     
  2. jcsd
  3. Oct 9, 2015 #2

    jbriggs444

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    This first question is easy. No, there does not have to be a coordinate system. A coordinate system is a human invention. It is purely a way of organizing and thinking about the physical situation. One can do physics without coordinates.
    This seems to confuse the field's potential at a point with the direction of the field at a point. A proper understanding of this involves the concepts of a "vector field", a "scalar field", "gradient" and "potential". That is vector calculus at its simplest level.

    The electric field (as a vector field) has both magnitude and direction. At every point in space there is an electric field strength and a direction in which that field points. The electric field (at least in the static cases we are considering) is "conservative". That means that along with this vector field is a potential field. The potential field is a scalar field. At every point in space there is an electric potential. This is a numeric value. It has no direction.

    Starting with the vector field, one can obtain the potential field by defining a point of zero potential and taking a "path integral" of the vector field to a chosen point. The computed path integral will be a scalar result. If the field is conservative, any path you choose will yield the same result. This result is the value of the potential field at that point.

    Starting with the potential field, one can obtain the vector field by evaluating its gradient. If we are using coordinates, the gradient is defined by taking partial derivatives of the potential field in each of the three coordinate directions and treating these results as a the components of a three dimensional vector.

    Now, back to where we started... The direction of the electric field as a vector field is opposite on opposite sides of a charge. The sign of the electric field as a potential field is the same on opposite sides of a charge.

    Unfortunately that means that the rest of your post went off into the weeds.

    If we (arbitrarily) designate the scalar field so that it is zero at infinity and if we are dealing with point charges then it is necessary and sufficient that there be a positive and a negative charge somewhere. If all the charges are positive then the potential is positive everywhere. If all the charges are negative then the potential is negative everywhere. If at least one charge is negative then it creates a pole in the field where potential becomes infinitely negative. If at least one charge is positive then it creates a pole where the potential becomes infinitely positive. The field is continuous. By the intermediate value theorem there will be a point (actually many points if we are working in three dimensions) where the potential is zero.
     
  4. Oct 9, 2015 #3
    The potential of a point charge [itex]q[/itex] at a point [itex]\bf{x}_0[/itex] is given by [itex] V(\bf{x})=\frac{q}{4 \pi \epsilon_0 r} [/itex] where [itex]r=|\bf{x}-\bf{x}_0|[/itex] (assuming the standard case where we take potential vanishing at infinity). Keep in mind that for two positive charges, this is always positive (since [itex]r[/itex] is always positive) so the total potential can never be 0 at a finite point. Similarly for two negative charges. The only way to solve your problem is with both positive and negative charges.

    Keep in mind that such a point is unlikely to have physical significance, since we can always add a constant to the potential without changing the dynamics.
     
  5. Oct 9, 2015 #4
    Why"r "can not be negative?
     
  6. Oct 9, 2015 #5

    jbriggs444

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    ##r = |x - x_0|## per the definition from Jack Davies. How can it be negative?
     
  7. Oct 9, 2015 #6
    A neutral point is not a point at which the potential is equal to zero. The potential is defined only with respect to an arbitrary constant, so a value of zero for the potential is meaningless.

    A neutral point is one at which, if a test charge is placed there, there will be no net force on the test charge. Thus, at such a point, the electric field intensity is zero and the gradient of the potential is equal to zero. In Cartesian coordinates, the gradient of the potential is equal to zero if the partial derivative of the potential with respect to all three of the coordinate directions is equal to zero.

    Chet
     
  8. Oct 10, 2015 #7
    What if the question says assume the potential at infinity to be zero.
     
  9. Oct 10, 2015 #8
    Then that establishes the value of your arbitrary constant.

    Chet
     
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