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Force on electric diploe in nonuniform electric field 
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#1
Jun2714, 02:10 AM

P: 176

I couldn't understand why there is,
∂E[itex]_{y}[/itex] and ∂E[itex]_{z}[/itex] term in the equation, for the xcomponent of the force on dipole, 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 xcomponent. I understand, the net force on the dipole, in an nonuniform 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 dipole are not the same in an nonuniform field. And therefore, there would be a net force on the dipole. 


#2
Jun2714, 03:12 AM

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What are the delta terms in the equation for?



#3
Jun2714, 03:35 AM

P: 4

can't get through what you have written. Bhai apka Post kia hua acha say nahe sam j



#4
Jun2714, 04:49 AM

P: 176

Force on electric diploe in nonuniform electric field
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).
I understand, force experienced upon xcomponent 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 xcomponent of the q charge= qE[itex]_{x}[/itex] and hence the net force on xcomponent of the dipole = 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] 


#5
Jun2714, 05:04 AM

P: 4

Hi you Indian ?



#6
Jun2814, 12:00 AM

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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. 


#7
Jun2814, 12:08 AM

P: 176

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. 


#8
Jun2814, 12:59 AM

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Sure ... the x component of the electric field can depend on y and z
$$\vec E = E_x(x,y,z)\hat \imath + E_y(x,y,z)\hat \jmath +E_z(x,y,z) \hat k\\$$ 


#9
Jun2814, 02:30 AM

P: 176

So, obviously ∂E[itex]_{x}[/itex]/∂y i.e rate of change of xcomponent on yaxis should be zero. Where am I wrong. 


#10
Jun2814, 02:41 AM

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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. 


#11
Jul1114, 08:43 PM

P: 176

For e.g a particle accelerating in the xy plane. If I'm right, its instantaneous velocity along xaxis 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 xaxis ∂V[itex]_{x}[/itex]/∂x = acceleration on the xaxis 


#12
Jul1214, 01:58 AM

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Didn;t you say earlier that the deltax deltay etc were related to the separation of the charges in the dipole.



#13
Jul1214, 05:36 AM

P: 176

Why the xcomponent depend on z or y axis? That is, once we know the rate of change along xaxis 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. 


#14
Jul1214, 08:19 AM

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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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