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


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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