Say I have 4 equal point charges in corners of a square. The electric field in the middle is 0. Now I brought another same charge from infinite hight to the middle of the square. I know the potential in the middle is not 0. The charge needs to move because there is potential difference (assuming 0 potential at infinity) But the field is 0 so it won't move. Then what is this situation? What does it mean in terms of energy I gave to the charge?
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2$\begingroup$ Why do you think it has to move? Just because something has potential energy doesn't mean it has to move. A ball sitting perfectly on top of a hill won't roll down unless pushed slightly. $\endgroup$– TriatticusCommented Feb 2 at 20:09
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2$\begingroup$ What do you mean by "has potential"? I can make the numerical value of the potential of at any point be +1,000,000 V or -1,000,000 V (or 0 V) just by choosing a different reference for my potential scale. $\endgroup$– The PhotonCommented Feb 2 at 20:27
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$\begingroup$ @ThePhoton at then end, OP fixed $V=0$ at infinity, also known as the physics-homework-gauge. $\endgroup$– JEBCommented Feb 2 at 21:54
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$\begingroup$ @JEB, but they should still think about what the arbitrariness of the reference potential means for the answer of their question. $\endgroup$– The PhotonCommented Feb 3 at 3:29
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2 Answers
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What you describe is an unstable equilibrium. A charge exactly at that point will feel no force. But any slight deviation in position from the exact center will produce a small force that accelerates the charge away from the equilibrium point.
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The absolute value of the potential is irrelevant. It's its derivative that matters.
Let $s$ be the side length of the square and $r=s\sqrt{2}/2.$
By symmetry on the z-axis, the electric field is
$\vec{E}=\frac{4ez\hat{k}}{4\pi \epsilon_0(r^2+z^2)^{3/2}}$
$V=\int_{0}^\infty\frac{2e^2(2z)\hat{k}\cdot dz \hat{k}}{4\pi \epsilon_0(r^2+z^2)^{3/2}}=\frac{-4e^2}{4\pi \epsilon_0}\frac{1}{\sqrt{r^2+z^2}}|_0^\infty=\frac{4e^2}{4\pi\epsilon_0r}$
What matters in terms of the field is local potential difference. At the center of the square $-\nabla V=0$ even though $V\ne 0$.
The absence of a field means absence of a force, that in turn means absence of a change momentum. So whatever momentum the particle has when it reaches the center, it maintains, and so will continue to pass through the center into a space with non-zer field. You will only have stability if the particle is at the center with no momentum relative tot he corner charges.
In practice, this is not possible given quantum mechanical effects. Either the particle isn't exactly in the center or it doesn't have exactly 0 momentum. In either case it's in a region of non-zero field or guaranteed to leave the one point with zero field.
