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What is the physical significance of such a point where electric field is non-zero but electric potential is zero? I mean, how can we understand this concept without mathematics?

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Strictly speaking potential is relative, so it's only the change in potential (and not the value itself) that matters – but let's assume that potential is set to zero at infinity.

What zero potential means, roughly, is that the charges in your system have cancelled out. For example exactly half way (or otherwise equidistant from them) between two equal and oppositely charged point charges, potential is zero.

If you move a particle between any two points of equal potential (zero or not) it doesn't cost any energy. So if you have a point with zero potential you can place a new particle there from outside the system (i.e. from infinity) for free.

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  • $\begingroup$ thank you for clarifying that up, so what is the point of the all hailed Laplacian and Poisson equation solution if we cant get E at all. $\endgroup$ – James Guana Jan 6 '16 at 22:48
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The point-values of the potential are not observable. You can only observe differences in the potential (for example by letting a charge move between two points and measuring the work done on it). Hence the physical significance of a point where the electric field is non-zero, is that the electric field is non-zero, and that the potential is zero there has no physical significance, because the point-values of the potential are not unique. You can for example always add a constant to the potential and have the same electric field, so that the potential is zero at a point can't have physical significance.

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I think answer to this is the Earth. Earth has an electric field but it doesn't have any potential.If it would any currents would have been produced.

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  • $\begingroup$ I think the near vacuum of the interplanetary space isn't a very good conductor. $\endgroup$ – peterh says reinstate Monica Aug 20 '17 at 12:38
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The electric field in terms of the potential is $\vec{E} = -\nabla V = -\frac{\partial V}{\partial x}\hat{x}-\frac{\partial V}{\partial y}\hat{y}-\frac{\partial V}{\partial z}\hat{z}$. So you see, the electric field only corresponds to some kind of differential of the potential i.e. how the potential changes in a particular direction.

It does not depend on the value of the potential itself. To illustrate this with a crude example, suppose the $x$ component of electric field at a point has a value 5 N/C. This could've been created by a potential difference of 0.005 over a distance of 0.001 i.e. $E_x \approx \frac{\Delta V}{\Delta x} \implies 5=\frac{0.005}{0.001}$. Note that $\Delta V = 0.005$. There is no mention of what $V$ value is at the two points. You can take the values to be $V_1 = 123456$ and $V_2 = 123456.005$ or just $V_1 = 0$, $V_2 = 0.005$. Now you see that the specification of the value of V is completely up to you. It is only the gradient of V that matters for the electric field.

I'll provide a cruder analogy, but this should help you understand what I've told in the previous paragraph. Imagine you're trying to measure the length of a pen. Let the length of the pen be 5 inches, say. When you measure it with the help of a ruler, it doesn't matter if one end of the pen coincides with the zero of the ruler or not. Hence the measurements 0 and 5 is equivalent to 3 and 8 , 4.65 and 9.65 and so on. All that matters is the difference between the measurements at the end points. As a matter of convenience, you could set one end to be 0 so that the readout on the other end corresponds to the length of pen. You should keep in mind that the zero has no significance and is only a matter of convenience in calculation.

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