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I came across an interesting yet perplexing statement in my physics textbook:

However, we hasten to add that electric fields produced by static electric charges have properties different from those produced by time-varying magnetic fields.

My questions are, simply put: What exactly are these properties? And why do they differ?

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The only difference is in the structure of the fields:

  • The field contribution from static charges are non-rotational (curl-free) but divergent.

  • The field contributions from time-varying magnetic fields are divergence-less but have non-zero curl.

You can read this right out of the Gauss and Faraday's Laws in their differential form: \begin{align*} \mathbf{\nabla \cdot E} &= \frac{\rho}{\epsilon_0} \\ \\ \mathbf{\nabla \times E} &= - \frac{\partial \mathbf{B}}{\partial t} \;. \end{align*} The operators on the LHSs are called "divergence" and "curl" (read top to bottom).


I hasten to add that these are exactly the same kind of stuff, which is why I've written "contributions from" rather than "fields produced by" as if the fields might be of different kinds. In the presence of both static charges and varying magnetic field the total electric field will show both divergence and curl.

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  • $\begingroup$ Got it... it's clear now. Danke. $\endgroup$ – Kugelblitz Sep 18 '16 at 3:46
  • $\begingroup$ You beat me @dmckee ;/ $\endgroup$ – user36790 Sep 18 '16 at 3:52
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$$\begin{array}{|c|c|} \hline\textrm{True in Statics} &\textrm{True in General}\\ \hline \mathbf F ~= \dfrac1{4\pi\varepsilon_o}~ \dfrac{q_1q_2}{r^2}~\mathbf{\hat r} & \mathbf F= q(\mathbf E+ \mathbf v\times \mathbf B)\\ \hline \nabla \cdot \mathbf E = \dfrac{\rho}{\varepsilon_0} & \nabla \cdot \mathbf E = \dfrac{\rho}{\varepsilon_0} \\ \hline \nabla\times \mathbf E= \mathbf 0& \nabla \times \mathbf E = ~-\partial_t\mathbf B\\ \hline \mathbf E= -\nabla\varphi & \mathbf E= -\nabla \varphi - \partial_t \mathbf A\\ \hline \nabla^2\varphi = -\dfrac{\rho}{\varepsilon_0} & \nabla^2\varphi -\dfrac1 {c^2}\partial^2_t\varphi = -\dfrac\rho{\varepsilon_0}\\ \hline \varphi(\mathbf 1)= \dfrac1{4\pi\varepsilon_0}\displaystyle\int\dfrac{\rho(\mathbf 2)}{r_{12}}~\mathrm dV_2 & \varphi(\mathbf 1, t)= \dfrac{1}{4\pi\varepsilon_0}\displaystyle \int \dfrac{\rho(\mathbf 2, t^\prime)}{r_{12}}~\mathrm dV_2\\\hline U= \dfrac12 \left(\displaystyle\int \rho\varphi~\mathrm dV \right) & U= \dfrac{\varepsilon_0}2 \left(\displaystyle\int \mathbf E\cdot \mathbf E~\mathrm dV + {c^2}\displaystyle\int \mathbf B\cdot \mathbf B~\mathrm dV\right)\\ \hline\end{array}$$

where

\begin{align}\mathbf F& \equiv \textrm{Total force},\\ \mathbf E& \equiv \textrm{Electric field},\\ \mathbf B &\equiv \textrm{Magnetic field},\\ \nabla &\equiv \textrm{del operator},\\ \rho &\equiv\textrm{Charge-density},\\ \varphi &\equiv \textrm{Scalar-potential},\\ \mathbf A &\equiv \textrm{Vector-potential},\\ \partial_t &\equiv \dfrac{\partial}{\partial t},\\ \nabla^2 &\equiv \textrm{Lapalacian operator},\\ \partial^2_t &\equiv \dfrac{\partial^2}{\partial t^2},\\ U &\equiv \textrm{Potential-energy}, \\ t^\prime &\equiv \textrm{retarded-time} = t-\dfrac{r_{12}}c\;. \end{align}

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  • $\begingroup$ How do you have $\mathbf{F} = \mathbf{E}$? $\endgroup$ – Mehrdad Sep 22 '16 at 10:35
  • $\begingroup$ @Mehrdad: Electrostatics.... $\endgroup$ – user36790 Sep 22 '16 at 10:52
  • $\begingroup$ Electrostatics can make force and field have the same units?? $\endgroup$ – Mehrdad Sep 22 '16 at 16:10
  • $\begingroup$ @Mehrdad: Sorry, lack of attentiveness; would fix this. Thanks for pointing out that. $\endgroup$ – user36790 Sep 22 '16 at 16:12
  • $\begingroup$ Mafia, I downvoted this because it gives a table and a list of terms, but it doesn't explain anything. $\endgroup$ – John Duffield Oct 18 '16 at 12:57

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