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I have a doubt... Electric field is the negative gradient of potential ... But curl of vector functions that are a gradient of a scalar function is the zero vector. Then how come Maxwell stated that $\text{curl} \mathbf{E} = -d\mathbf{B}/dt$?

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Electric field is the negative gradient of potential

This is not true. The fundamental law is Faraday's law, which tells us that $\nabla \times \mathbf E = -\frac{\partial}{\partial t} \mathbf B$. As a result, $\mathbf E$ generally cannot be written as the gradient of a scalar potential, for precisely the reason you say. However, if $\frac{\partial}{\partial t}\mathbf B = 0$, then it may be possible to write $\mathbf E = -\nabla \varphi$ for some scalar function $\varphi$.

Assuming that $\mathbf E$ is smoothly defined over all of $\mathbb R^3$, then this turns out to be true. If you are restricting to a region $R\subseteq \mathbb R^3$, then you are guaranteed to be able to find a single function $\varphi$ such that $\mathbf E = -\nabla \varphi$ if $R$ is simply-connected; if it isn't, then you may not be able to find a single $\varphi$ to do the job on all of $R$.

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  • $\begingroup$ so you mean this relation between electric field and potential is only true when dB/dt is zero? $\endgroup$
    – Varsha
    Dec 2, 2022 at 16:30
  • $\begingroup$ @Varsha Yes. More generally, we can write $\mathbf B = \nabla \times \mathbf A$ for a so-called vector potential $\mathbf A$; if we do this, then we find that $\nabla \times \left( \mathbf E +\frac{\partial}{\partial t} \mathbf A\right) = 0$ and so, assuming the space is simply connected, we can write $\mathbf E +\frac{\partial}{\partial t} \mathbf A = -\nabla \varphi \implies \mathbf E = -\nabla \varphi - \frac{\partial}{\partial t}\mathbf A$. If $\mathbf B$ is constant in time, it just turns out we don't need the $\mathbf A$. $\endgroup$
    – J. Murray
    Dec 2, 2022 at 16:38
  • $\begingroup$ @J.Murray Curl of something is zero not necessarily means it is gradient of scalar potential. Maybe it is homogeneous and isotropic in space, maybe not a vector or least if it is sum then sum equals to zero. $\endgroup$ Dec 2, 2022 at 18:23
  • $\begingroup$ @NeilLibertine As I stated in my answer, if $\mathbf E$ is differentiable on some domain $R$ and $\nabla \times \mathbf E = 0$, then we are guaranteed that $\mathbf E$ can be written as the gradient of a scalar field if $R$ is simply connected. I have no idea what you're trying to say in the second part of your comment. $\endgroup$
    – J. Murray
    Dec 2, 2022 at 19:32
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That is because electric field is given as,$$\mathbf E=-\nabla\phi-\frac{\partial\mathbf A}{\partial t}$$and magnetic field is given as,$$\mathbf B=\mathbf\nabla\times\mathbf A$$If one see then finds that physicists rely on mathematics from the days of eighteenth century or before. It was found that electric field is source dependent, so have divergence, and which can be expressed as gradient also. Later work of faraday with magnetic induction, it seems that electric field along loop due to change in magnetic field. So they modified it with above expression using vector potential.

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The electrostatic field is the gradient of the electric potential. However electric fields in general do not follow this relationship.

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  • $\begingroup$ Why do electric fields not follow this relationship? $\endgroup$
    – Varsha
    Dec 2, 2022 at 16:26
  • $\begingroup$ @Varsha At a high level, it's a mathematical description of a result obtained from experiment. You can also obtain Maxwell's equations from a limit of quantum electrodynamics, though. $\endgroup$
    – agaminon
    Dec 2, 2022 at 16:38
  • $\begingroup$ @Varsha We can't tell you "why" without resorting to circular reasoning. The theory was constructed to match the results of experiments like micro.magnet.fsu.edu/electromag/java/faraday2/index.html. Not a difficult experiment, you should try it. $\endgroup$
    – John Doty
    Dec 2, 2022 at 17:54

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