How can one understand intuitively that without magnetic field, the potential can still exist? Also $\nabla^2 \Phi=−\rho/\epsilon_0$. If charge density is $0$,$\Phi$ is non zero. How can potential still exist without charge?
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1Probably useful: https://physics.stackexchange.com/q/53020/25301 – Kyle Kanos Oct 06 '23 at 01:55
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9Your first question (the question in the title of the post), is asking no more and no less than: If the derivative of a function is zero, why isn't the function also zero? The function is a constant, since the derivative of a constant is zero. This is just how math works. – hft Oct 06 '23 at 02:29
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1Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking. – Community Oct 06 '23 at 03:12
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1Consider the case where there is a region with some mass (or charge) and the rest is empty. Now, you expect to have a gravitational (or electric) field everywhere, even in the empty region: we observe that this kind of fields can propagate in vacuum once they are somehow generated somewhere. So, the fact that $\nabla^2 \phi=0$ does not imply necessarily $\phi=0$ it's a mathematical property of the Laplacian operator that "fits" what we observe in nature, and so that's why it is used. (It fits also many other facts). – Quillo Oct 06 '23 at 06:17
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1The prime example of a vector potential $A$ that leads to a zero magnetic field $B$ is $$ A=\frac{1}{x^2+y^2}\begin{pmatrix}-y\x\0\end{pmatrix},. $$ You may want to get familiar with the Aharonov-Bohm-effect. – Kurt G. Oct 06 '23 at 15:41
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If $\mathbf{A}=\nabla\phi$ where $\phi$ is an arbitrary scalar field, then we have $$\nabla\times\mathbf{B}=\nabla\times\nabla\phi=0.$$ Since $\phi$ is arbitrary, it is easy to choose a $\phi$ that makes $\mathbf{A}\neq0$. $\mathbf{B}$ is physical. $\mathbf{A}$ is simply a mathematical tool to make calculation about magnetic field easier. It does not have any physical meaning in itself.
Regarding electric potential, same thing goes. It is merely a mathematical tool to simplify equations in certain cases.
Using these auxiliary fields is similar to the following use of $a$:
Suppose we have an equation $$x+f(x)+g(x)=\left(\frac{c_1+c_2}{c_3}+c_4\right)^{c_5}+c_6c_7$$ where $c_i,(i=1,2,...,7)$ are constants. We can simplify this equation by defining the auxiliary parameter $$a\equiv\left(\frac{c_1+c_2}{c_3}+c_4\right)^{c_5}+c_6c_7$$ The equation then becomes $$x+f(x)+g(x)=a$$ In the consequent calculations, you need not to deal with the large number of $c_i$ that cause inconvenience.
Yufei
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As it’s currently written, your answer is unclear. Please [edit] to add additional details that will help others understand how this addresses the question asked. You can find more information on how to write good answers in the help center. – Community Oct 07 '23 at 06:00