What is the physical reason a $+5V$ equipotential coutour cannot intersect a $-5V$ equipotential coutour?

Question

Now I've been told that equipotential contours with different values can never intersect. That is, if one level is 5V and one is -5V, they can't intersect. This make sense to me mathematically (one contour is above the other), but how do you explain that this is true in terms of energy? In other words, how is energy even related to this problem? Because I don't see the relationship at all.

Answer 1

While it may be possible to derive a violation of energy conservation due to intersecting equipotentials, there is a much more intuitive and in my opinion a more fundamental reason that equipotentials cannot intersect:

Potential is a single-valued function.

A good analogy for potential in this case is a map of the ground elevation of the earth; a Topographic Map. Like potential, elevation can only take one value at each point in the domain over which it is defined. Equipotential lines are sets of points in that domain which all have a particular value of potential. In a topographical map, there are contour lines, which represent sets of points that have the same height.

Now, lets say you are looking at a contour line on a map, and that contour represents the points on the surface with an elevation of 100 meters above sea level. If you go to that place on the map, and walk along that line, you will stay at 100m. Now, look at the contour on the map next to the one you are on; maybe it marks the 110m contour. This means that every point on that line is at 110m above sea level.

Now, can you see what it would mean if that contour line intersected the one you are on? Would you be at 100m or 110m? If your friend was following the other contour, happily staying at 110m above sea level, and then you met at the spot where the lines intersect, would you disagree over your altitude? Of course not! This situation is impossible.

To summarize briefly, contours cannot intersect because they represent regions of constant value. If two contours represent different values, then they cannot intersect because each point may only have one value.

Answer 2

There is a kind of energy associated with an electrically charged particle placed in electric field which is called electrostatic potential energy. It is the product of the electric potential V and the electric charge q of the particle:

\begin{equation} E_p=qV \end{equation}

Note that the description of the electric processes does not change if you add the same constant amount to the potential in all places in space^*. Thus, what really matters are the differences in potential between different points and the resulting gain or loss of energy a charged particle experiences as it moves from one place in the electric field to another. This is why it only makes sense to speak of potential energy with reference to some reference point which is chosen to have potential V=0. In order to determine the amount of potential electric energy gained or lost by an electrically charged particle as it moves from A to B one needs to multiply the charge of the particle by the difference of the electric potential between the two points:

\begin{equation} E_p=q(V_A-V_B) \end{equation}

Note an interesting thing here^*: the amount of electric potential energy gained or lost by the particle doesn't depend on the path that the particle took through space, but only on the potential difference between the starting and final points.

Now, answering your question directly: if the equipotential contours were to intersect, both the value of the electric potential at the intersection point and the electric potential energy of a particle placed in that point would be undefined. You could use the value of the potential from any of the intersecting contours to arrive at different values of the electric potential energy.

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Aside: there is a unit of energy which derives from the relationship between charge, potential and energy described above. It's called electronvolt: 1eV is the amount of electrostatic potential energy that a particle with electric charge equal to the charge of electron gains as it moves across electric potential difference of 1V.

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^*The curious properties of static electric field alluded to above are formalized using mathematical notion of conservative vector field (of which electric field produced by a group of static charges is an example).

Answer 3

The existing answers by Adam and Colin are sufficient, but I'd like to encourage you to consider the physics of trying to make the lines come "almost together".

If you are just studying this subject for the first time you may not yet have encountered all the concepts that follow. Don't worry about that, just put it on the back burner.

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An exercise that might be instructive is to compute the electric field strength in a region where a +5 Volt and -5 Volt surface are brought very close together (say, 1 mm). How does that compare to the breakdown voltage of various materials. What if you reduce the separation to a 0.1 mm? To 0.01 mm? To a micrometer? The form of the answer is $$ E \propto \frac{V_2 - V_1}{d}$$ for $d$ the separation distance. Notice that the field strength would grow without bound if you allowed the separation to go to zero: always a sign that you're in dicey territory.

These issue become critical in the design of compact capacitors.

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Electric fields also carry energy. So, how much energy per unit volume is required to maintain two different potential surfaces very close together? (You'll want to make some simplifying assumptions about the geometry if you're going to calculate exactly, or go go with the form of the equations. The form of the answer here is $$U \propto E^2 \propto \frac{(\Delta V)^2}{d^2}$$ which may be what was meant by causing problems related to energy if you try to make them come all the way together: this diverges even faster than before.