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In my electrodynamics text book while deriving Force on a dipoe in a nonuniform electric field it says
                  F = q$E_+$ - q$E_-$
     implying this         $q(\bigtriangleup E_x\hat{x}+\bigtriangleup E_y\hat{y} + \bigtriangleup E_z\hat{z})$
till this i get how we got here but then it say "taking infinitesimal distance, as would be for an ideal dipole , we can write change in field as the dot product of the gradient of each component with nfinitesimal displacement dl " changing the above equation to this
                  $\bigtriangleup E_x = \bigtriangledown E_x.dl$                     $F = (\textbf{p} .\bigtriangledown)\textbf{E}$
How can change in Electric field equate to this and why it was assumed that dipole is infinitely small

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Yup faced the same confusion but see it is very logical and mathematical while it seems totally bizarre ; let me try to explain in a concise manner. enter image description here let's consider a one dimensional case and for simplicity assume the dipole to be placed along the axis of variation of the field. Then the the force on the charges ( i.e. , $qE$ ) can be written as $ F = qE(x) - qE(x-l) $ where $ l $ is the length of the dipole multiplying both the Nʳ and the Dʳ by $ l $ we can notice a form of the first principle of derivatives if $ l $ tends to zero ! from here we get that $ F = p \tfrac {dE}{dx} $ now assuming a multidimensional case we can use the "principle of superposition" to extend the explanation beyond 1-D. In more than one dimension the only thing that would change is that the normal multiplication would be changed into dot product ( as orientation matters ) and simple differential would become a gradient ( calculated easily using nabla operator )

Feel free to ask more doubts suggest edits and wrong reasoning in the post I would be happy to learn more !

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  • $\begingroup$ Ok so $(p. \bigtriangledown)E$ is only to differentiate every component of E in x,y,z direction in 3d $\endgroup$ – tanuj23199 Jul 30 '19 at 6:37
  • $\begingroup$ It's $ p. (\nabla E) $ yup what you say is right though it is an extension of derivatives $\endgroup$ – Aditya Garg Jul 30 '19 at 8:29

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