Force on electric diploe in non-uniform electric field

  1. Jun 27, 2014 #1
    I couldn't understand why there is,
    ∂E[itex]_{y}[/itex] and ∂E[itex]_{z}[/itex] term in the equation,
    for the x-component of the force on di-pole,

    F[itex]_{x}[/itex] = q [ E[itex]_{x}[/itex] + ∂E[itex]_{x}[/itex]/∂x δx + ∂E[itex]_{y}[/itex]/∂y δy + ∂E[itex]_{z}[/itex]/∂z δz ] - qE[itex]_{x}[/itex]

    Isn't both ∂E[itex]_{y}[/itex] and ∂E[itex]_{z}[/itex] term, should be zero along the x-component.

    I understand, the net force on the di-pole, in an non-uniform electric field, should be,
    F[itex]_{x}[/itex] = q [ E[itex]_{x}[/itex] + ∂E[itex]_{x}[/itex]/∂x δx] - qE[itex]_{x}[/itex]

    Since the force on the ends of a di-pole are not the same in an non-uniform field. And therefore, there would be a net force on the di-pole.
     
  2. jcsd
  3. Jun 27, 2014 #2

    Simon Bridge

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    What are the delta terms in the equation for?
     
  4. Jun 27, 2014 #3
    can't get through what you have written. Bhai apka Post kia hua acha say nahe sam j
     
  5. Jun 27, 2014 #4
    Its given E[itex]_{x}[/itex], E[itex]_{y}[/itex] and E[itex]_{z}[/itex] are three rectangular component of field strength E at the origin where the charge -q of the dipole is situated and the charge +q is situated at (δx, δy, δz).

    The delta terms must be the rate of change of the field.


    I understand, force experienced upon x-component of the +q charge
    = q [ E[itex]_{x}[/itex] + ∂E[itex]_{x}[/itex]/∂x δx + ∂E[itex]_{y}[/itex]/∂y δy + ∂E[itex]_{z}[/itex]/∂z δz ] and,

    force experienced upon x-component of the -q charge= qE[itex]_{x}[/itex]

    and hence the net force on x-component of the di-pole =

    F[itex]_{x}[/itex] = q [ E[itex]_{x}[/itex] + ∂E[itex]_{x}[/itex]/∂x δx + ∂E[itex]_{y}[/itex]/∂y δy + ∂E[itex]_{z}[/itex]/∂z δz ] - qE[itex]_{x}[/itex]
     
  6. Jun 27, 2014 #5
    Hi you Indian ?
     
  7. Jun 28, 2014 #6

    Simon Bridge

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    NO - the delta terms are the position of one end of the dipole with respect to the other one.
    The partials attached to the delta terms are the rate of change of E with the direction.

    Since the dipole can be tilted in the y or z direction, the gradient of E in that direction must count.
     
  8. Jun 28, 2014 #7
    Apologize, there is no ∂E[itex]_{y}[/itex] and ∂E[itex]_{z}[/itex] in the equation, so the equation is F[itex]_{x}[/itex] = q [ E[itex]_{x}[/itex] + ∂E[itex]_{x}[/itex]/∂x δx + ∂E[itex]_{x}[/itex]/∂y δy + ∂E[itex]_{x}[/itex]/∂z δz ] - qE[itex]_{x}[/itex].

    Still, can there be term ∂E[itex]_{x}[/itex] wrt ∂y and ∂z. I understand it would be zero wrt ∂y and ∂z.
     
  9. Jun 28, 2014 #8

    Simon Bridge

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    Sure ... the x component of the electric field can depend on y and z
    $$\vec E = E_x(x,y,z)\hat \imath + E_y(x,y,z)\hat \jmath +E_z(x,y,z) \hat k\\$$
     
  10. Jun 28, 2014 #9
    We have only E[itex]_{x}[/itex] component along x-axis, similarly E[itex]_{y}[/itex] and E[itex]_{z}[/itex].

    So, obviously ∂E[itex]_{x}[/itex]/∂y i.e rate of change of x-component on y-axis should be zero.

    Where am I wrong.
     
  11. Jun 28, 2014 #10

    Simon Bridge

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    ... that's what I wrote - notice that each component of the electric field is a function of position?

    If the x component electric field does not depend on z or y then the gradient in those directions will be zero.
    The equation you wrote does not make that assumption.
    The equation is explicitly for the situation that the electric field varies with position.
     
  12. Jul 11, 2014 #11
    I still am missing something.

    For e.g a particle accelerating in the x-y plane.

    If I'm right, its instantaneous velocity along x-axis after an interval δx will be, ∂V[itex]_{x}[/itex]/∂x δx, and we wouldn't need ∂V[itex]_{x}[/itex]/∂y δy.

    V[itex]_{x}[/itex] = instantaneous velocity on the x-axis
    ∂V[itex]_{x}[/itex]/∂x = acceleration on the x-axis
     
  13. Jul 12, 2014 #12

    Simon Bridge

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    Didn;t you say earlier that the delta-x delta-y etc were related to the separation of the charges in the dipole.
     
  14. Jul 12, 2014 #13
    I think this is what I'm still not getting. Can you please make it simple.

    Why the x-component depend on z or y axis? That is, once we know the rate of change along x-axis is ∂E[itex]_{x}[/itex]/∂x, we don't need z and y component.

    Only ∂E[itex]_{x}[/itex]/∂x is required to know the value of E[itex]_{x}[/itex] at δx.

    Yes, one end of the dipole is at origin and the other charge +q is situated at (δx, δy, δz).
     
  15. Jul 12, 2014 #14

    Simon Bridge

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    ... that would happen if the field is not uniform.
    i.e. For a point charge, the electric field is $$\vec E = \frac{kQ}{r^3}\vec r$$ ... find ##E_x##

    Remember:
    The electric and magnetic fields are vectors.

    So ##\vec E = E_x(x,y,z)\hat\imath + E_y(x,y,z)\hat\jmath + E_z(x,y,z)\hat k##

    ##E_x## is not the value of ##\vec E## along the ##x## axis, it is the component of ##\vec E## at point ##\vec r = (x,y,z)## that points in the +x direction.
     
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