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Electrostatic fields- What defines an electrostatic field?

  1. Apr 4, 2014 #1
    I have a question about electrostatic fields!

    I was given these two equations


    E= k [xy xˆ + 2yz yˆ + 3xz zˆ]

    E= k[y^2 xˆ+ (2xy+ z^2)+ 2yzzˆ]

    And asked to determine if either could be an electrostatic field, I'm not asking how to solve this problem but I'm not sure what defines an electrostatic field!

    I have heard vector and scalar fields being mentioned and I know that curl and divergence pertain to vector fields and gradient to scalar fields:)

    My initial thought was to calculate divergence and curl and draw conclusions from there?

    I've combed through passages on my textbook and notes and can't find an explicit answer as to what defines an electrostatic field (other than static point charges)!


    Any help would be great :)
     
  2. jcsd
  3. Apr 4, 2014 #2
    Oh I also missed out my "y" unit vector in the second expression, apologies
     
  4. Apr 4, 2014 #3
    Try applying Faraday's law.
     
  5. Apr 4, 2014 #4
    I thought that but it hasnt been mentioned in lectures so I was hesitant
     
  6. Apr 4, 2014 #5
    I'm assuming manipulation of the functions is where the answer is, so is there no relationship between curl, divergence and gradient and electrostatics?
     
  7. Apr 4, 2014 #6
    Just think of what a static field means. Think of what it implies about the direction of the field lines. If I have a stationary charge (static), what would I expect its divergence to look like? What about its curl?
     
  8. Apr 4, 2014 #7
    I would expect curl to be zero?

    And divergence to be positive or well rather the same in all directions
     
  9. Apr 4, 2014 #8
    Correct, now go one step further and think about what the divergence implies about the physical charge in space, and what you would expect to happen when you move the charge. Compare what you think to Maxwell's Equations (Gauss' Law and Faraday's Law)
     
  10. Apr 4, 2014 #9
    I know if the charge were moved the field would remain unchanged
     
  11. Apr 4, 2014 #10
    Are you sure about that?
     
  12. Apr 4, 2014 #11
    Not anymore haha!! So would if charges were moved the field could potentially develop a curl and differ in divergence
     
  13. Apr 4, 2014 #12
    Well the divergence would remain unchanged, but yes, a curl would develop. This curl is a time-varying magnetic field. Charges in motion produce time-varying magnetic fields!
     
  14. Apr 4, 2014 #13
    That makes complete sense!

    So to I would know if I had an electrostatic field if it had no curl, and if it's divergence was in the same direction for each component?

    Would gradient not then be in the radial direction of the field as well in this case?
     
  15. Apr 4, 2014 #14

    jtbell

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    Staff: Mentor

    This is Problem 2.20 in Griffiths, "Introduction to Electrodynamics" (3rd ed.).

    If you're using Griffiths, flip back to section 2.2.4, "The Curl of E". :wink:
     
  16. Apr 4, 2014 #15
    It just so happens I am using Griffiths haha!!

    Thanks a bunch friend! I appreciate it
     
  17. Apr 4, 2014 #16

    jtbell

    User Avatar

    Staff: Mentor

    Divergence (##\vec \nabla \cdot \vec E##) is a scalar. It doesn't have a direction. As for whether the divergence needs to be anything in particular, what is the divergence related to, via Gauss's Law?
     
  18. Apr 6, 2014 #17
    Divergence is related to Gauss law as


    Δ.E= ρ/εo
     
  19. Apr 6, 2014 #18
    Whoops that was supposed to be del, not delta
     
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