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Electric Potential of a Point Charge close to R=0

  1. Mar 19, 2016 #1
    1. The problem statement, all variables and given/known data

    What is a 3D representation of voltage using Kq/r assuming a positive point charge and what is the equation in cartesian and cylindrical form

    2. Relavent equations

    3. Attempt at solution

    I was trying to get a better understanding of Voltage, to really FEEL voltage... So I drew a graph of a 3D depiction of what I believed to be the representation of a +q point charge. It's pretty much a hill where the max is where the voltage is highest assuming the point charge is positive and levels off asymptotically at 0 as r increases.

    Is this a correct graph? Because if voltage increases as a radius decreases for a positive point charge, shouldn't the voltage move to infinity as you get closer and closer to the positive point charge?

    So the graph should really be a |1/x| with both sides approaching infinity instead of a |1/x| with a curved top with a maximum?

    Then I wanted to know the equation of the 3D surface generated. How do I convert V(r) = kq/r to a 3d surface?
    For cartesian for I'm guessing just Kq / root(x^2 + y^2)

    Are there any possible applications of understanding the 3D graph of voltage of a point charge?
  2. jcsd
  3. Mar 19, 2016 #2
    1. Yes the graph is assumed is correct with asymptote at x=0,y=0

    2. The graph will not have a curved top. It will approach to infinity at x=0,y=0
    3. As you said 3d equation seems obviously to be [tex] z = \frac{ kq}{\sqrt{ { x}^{2}+{ y}^{2}}} [/tex]

    4. A possible application would be finding out the distance of closest approach of a +ve charged particle approaching the +ve charge in question, assuming the positive charge in question is fixed. Consider a simulation where approaching charge is a particle (or a ball with very small radius ie almost zero moment of inertia) thrown in towards with tip of cliff (which is actually at infinity) with its velocity parallel to surface on which the ball is.

    electric potential analogy.png

    a) There will also be a direct analogy between velocity of particle in the simulation at any position x,y and velocity of approaching charge in original ques.

    b) There will be no direct analogy between time particle/ball takes to cover distance between x1,y1 to x2,y2 and time charged particle takes to cover distance between x1,y1 to x2,y2. This will be due to the fact that particle/ball will be travelling in z direction also and thus will take more time comparatively.

    c) If the approaching charged particle is not thrown exactly towards the fixed charged particle then this simulation will not work. And there are many reasons to it. First being the traversal in z direction of the ball/particle. Second the total force hill and gravity applies on the particle has component in z direction. Even in the ball case angular momentum about the center of hill is not conserved (whereas the angular momentum of approaching charged particle was conserved about the fixed charged particle in the original case)
  4. Mar 19, 2016 #3
    Good explanation. I most likely didn't completely understand all of it, but I got the main jist of 4a and above... the ball rolling towards the edge of the cliff makes sense but in this case it does not apply correct? Because if a positive q+ charge approaches the stationary +q charge, it is really going up a "hill" doing work to approach the stationary q+ point charge.. The only reason I'm saying that is because this graph assumes that the charge approaching is positive bc it's usually what most base equations and circumstances assume.

    How does the derivative of kq/r relate to the graph? dV/dr = -kq/r^2 or approx = -kq / (x^2 + y^2)
    Why does it seem that the slope is always negative for the graph which isn't true in this case?
    This would only be true if the derivative were taken of the absolute value of |kq/r|

    Why does the equipotential 3D graph we create represent the absolute value of kq/r where as kq/r itself doesn't matter?
    Is it because in polar form or cylindrical coordinates, r doesn't denote direction, just magnitude? So this means there must be a θ involved in the original kq/r equation correct?
  5. Mar 19, 2016 #4
    In my explanation, I clearly stated this assumption (ie moving charge is assumed positive).
    However if you want to draw the same analogy for a negative charge then just change z to -z ie replace cliff to something like a well.

    I am not sure what you mean here. But please check if you are not making the mistake of taking the direction of r wrong.
    when you say that dV/dr = -kq/r^2 then dir of r is assumed to be along the line joining the fixed charge to moving charge and in that case dV/dr and slope both are -ve. And the vice versa is also true.

    Didn't get you. Please elaborate.
  6. Mar 19, 2016 #5
    The graph of kq/r in the x-y plane lies in the 1st and 3rd quadrant. By making the equation |kq/r| It restricts the equation to the first and second quadrant.
    14.1.r.6.gif this is the equation 1/r, by making it |1/r| it would restrict the graph to y>0 or voltage > 0 because it is a positive point charge.

    I am asking, why does kq/r create a different graph where the absolute value of kq/r is what we really want? I believe it is because the radius value is a MAGNITUDE, there are no negative radius per se.

    The reason I wanted to take the derivative of |kq/r| was to find a relationship between the directional derivative of the graph and the voltage, possibly slope denoting direction or potential energy because putting in a +q or -q in the numerator would change the slope of the graph correct?

    so then the equation should really be kq/|r| in terms of the graph. I guess I'm confused finding the cylindrical equation for the graph.

    I got Du V(x,y) = kqx/(x2+y2)3/2*cosθ + kqy/(x2+y2)3/2*sinθ

    and dV/dr = kq/r * 1/|r|
    Last edited: Mar 19, 2016
  7. Mar 19, 2016 #6
    The whole problem lies when you say that potential is defined as kq/r (here r is distance) and tries to plot r on -ve x axis
    Lets be clear on what parameter you choose to represent on x-axis.

    a) If you represent distance on x-axis then curve will never lie in 2nd or 3rd quadrant as distance can never be -ve.

    b) If you represent position of point on x-axis then the curve we get is in 1st and 2nd quadrant.
    Assuming a simple case of 1d (ie plotting for all points lying on x-axis)
    Here y-axis represents potential of a point x-axis represents x co-ordinate of the point.

    Please feel free to ask for clarification where you have a doubt.
  8. Mar 19, 2016 #7
    I agree with your statement a) which is why I believed you must put kq/ |r| because the r represents the magnitude, there isn't a negative r direction which would cause the original kq/r equation to not work in 3d space...

    I think you are right

    I am still trying to figure out the relationship of the slope of the graph with voltage and potential energy.
    The derivative of kq/r is the rate of change of voltage with respect to r. If we place a point charge on those points, will the q sign alter the slope and direction of the point charge?
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