text: Homework Help: Voltage in the Middle of an Electric Dipole
  
  1. Apr 25, 2007
  
  kaotak
  
  Problem Diagram (Ignore the tildes, they're just placeholders):
  
  Below: An electric dipole
  ~~~~~~~y-axis~~~~~~~~~~~~~~~~~~
  ~~~~~~~|~~~~~~~~~~~~~~~~~~~~~
  ~~~~~~~|~~~~~~~~~~~~~~~~~~~~~
  ~~~~~~~|~~~~~~~~~~~~~~~~~~~~~
  ~~<---a---> <---a--->~~~~~~~~~~~~~~~
  +Q --------- X --------- -Q~~~-------------- x-axis
  
  Problem Statement: Find the voltage at X.
  
  My answer is that V_x = 0, since the potentials from each side of the dipole sum to zero. And I'm pretty sure this is right. But my question is... how does this fit in with the following definition of voltage:
  
  "The voltage at an arbitrary point P is the amount of work per unit charge it takes to move a test charge from infinity to P" (Physics for Scientists and Engineers)
  
  I see that if the test charge is approaching X from south of X, the work will be zero, since there is only a force in the east direction as the +Q and -Q cancel e/o out in the y-direction. Same thing if the test charge is approaching x from north of X.
  
  But what if the test charge is approaching X from west of X? east of X? How is the work zero? Won't the work be infinite from the west, assuming a positive test charge, because of the asymptotic behavior of the electric field along that line? Won't it be negatively infinite from the east, assuming a positive test charge?
  
  So, can someone explain how the work is zero coming from infinity west or east of X?
  
  I did think to myself that the work should be path-independent... but it's not... so how would you explain this?
  
  2. Apr 25, 2007
  
  mezarashi
  
  It is not infinity coming from either west or east. This is because the electric field due to these charges at infinity is zero! The electric field decays very quickly as you move away from these charges. If you do the calculus, you'll find that it takes pretty much negligible work to move any charges around at distances of large multiples of a, since E= kq/r^2 becomes very very small.
  
  Try doing the integration yourself for a single charge to convince yourself that no such infinity exists.
  
  3. Apr 25, 2007
  
  kaotak
  
  At infinity the electric field is zero. yes. But what about when the test charge is right ontop of +Q as shown in the diagram? And when it is right ontop of -Q as shown in the diagram?
  
  4. Apr 25, 2007
  
  mezarashi
  
  Well, in theory (according to these basic rules) if you have a two positive charges, it is impossible to get them "ontop" of one another, because it would require infinite force. The force of repulsion would be kq1q2/r^2. As r approaches zero, the force approaches infinity. Remember that we have estimated these charges to be "point charges", meaning that they are a singularity. This does not exist in the real world per se.
  
  What I can say is that, this is a model that we've created to help us solve problems, if you are looking at the boundary cases, the answers might become a bit dodgy.
metadata: {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}