when we say there is zero potential at the mid point of two opposite charges of equal magnitude, doesnt it mean that the work done in bringing a charge from infinity to that point is zero? iam not getting this point, why the potential energy is zero at this point.
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1$\begingroup$ it's like being in the valley between two equally sized hills $\endgroup$– pentaneCommented Sep 1, 2018 at 19:21
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2$\begingroup$ Probably more like mid point between hill (positive charge, positive potential) and valley (negative charge, negative potential). $\endgroup$– npojoCommented Sep 1, 2018 at 20:16
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4 Answers
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In this case there is one specific path all the way from infinity to the mid point, the midperpendicular between charges, which exhibits no forces parallel to the path.
Along this path, work obviously sums to zero.
Work along other paths would also be zero but would consist of positive work nullified by a balancing negative work.
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1$\begingroup$ Excellent answer. Also, because the field is conservative you are guaranteed that all paths to the same point will have the same value. That justifies the last paragraph. $\endgroup$– DaleCommented Sep 1, 2018 at 23:09
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The potential energy at any point in space is the sum of potential energies contributed by each charge.
Since the charges have equal magnitude and the distance from each to the mid point is the same, the magnitude of the potential energy contributed by each charge is the same, but the signs are opposite, so the net potential energy should be zero.
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$\begingroup$ @AaronStevens Thank you for pointing out that my original answer was based on the wrong understanding of the question, i.e., that the charges had the same sign. $\endgroup$– V.F.Commented Sep 1, 2018 at 20:27
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$\begingroup$ Of course! Always enjoy your answers. Some of the best ones here in my opinion. $\endgroup$ Commented Sep 1, 2018 at 21:09
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I think that the interesting thing is that to get a unit positive test charge from infinity to the mid point of two opposite charges of equal magnitude requires an external force to act on the test charge but in the process the external force does no work.
Suppose the two charges in question are at positions $C$ and $D$ and the unit positive test charge is moving along the path $AB$ where $B$ is the mid point between the two charges at $C$ and $D$.
The force on the test charge due to the positive charge at $C$ is $\vec F_{\rm ++}$ and this force has components $F_{\rm ++x}$ and $-F_{\rm ++y}$ in the $\hat x$ and $\hat y$ directions respectively.
The force on the test charge due to the negative charge at $D$ is $\vec F_{\rm +1}$ and this force has components $F_{\rm +-x}$ and $F_{\rm +-y}$ in the $\hat x$ and $\hat y$ directions respectively.
However the magnitudes of forces $\vec F_{\rm ++}$ and $\vec F_{\rm +-}$ and their equivalent components are all the same.
So the net force on the test charge is $$F_{\rm ++x} \,\hat x + F_{\rm +-x} \,\hat x + (-F_{\rm ++y}) \,\hat y + F_{\rm +-y} \,\hat y = 2F_{\rm ++x} \,\hat x$$
The external force on the test charge must be $-2F_{\rm ++x} \,\hat x$ whose direction is at right angles to the direction of the path $(\hat y)$ it has to take to reach position $B$, so that external force does no work.
Thus the potential at infinity and all points along line $AB$ is the same.
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It isn't zero potential, unless you decide that it is. All electrical potentials have an arbitrary additive constant. Potential differences are determined by physical conditions, and are measurable, but absolute potential values are only defined, or assigned, and not measured.
