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In schools we learn about electric field of point charge being negative gradient of potential. From this how can we generalize it for charge distributions? Any proof will be appreciated.

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  • $\begingroup$ About the negative: that is just a convention, we notice that in electrostatic $rot \bf{E} = 0$ and from that we deduce that the electrostatic field derive from a potential, we usually write it $\bf{E}=-\nabla \phi$ but nobody prevent me from defining $f=-\phi$ and from that writing $\bf{E}= \nabla $$f$calling $f$ the potential $\endgroup$ – Run like hell Jan 23 '17 at 8:20
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Thanks to the principle of superposition: you can imagine an arbitrary charge distribution as the superposition of infinitesimal point charges. $$ \rho (\textbf{x}) = \sum dq_i\, \delta(\textbf{x} - \textbf{x}_i) $$ Therefore the potential of the distribution is the sum (possibly the integral) of the potentials due to every single charge: $$ V(\textbf{x}) = \sum V_i(\textbf{x}) \,\,\,\left(= \frac1{4\pi\epsilon_0}\int\frac{\rho(\textbf{x}')}{|\textbf{x} - \textbf{x}'|} d^3x'\right) $$ Since the sum of the gradient is the gradient of the sum, $$ \textbf{E(x)} = \sum \textbf{E}_i(\textbf{x}) = \sum \left( - \nabla V_i(\textbf{x}) \right) = - \nabla \left(\sum V_i(\textbf{x})\right) = - \nabla V(\textbf{x}) $$ you have shown that the electric field of a charge ditribution is the negative gradient of its potential.

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