Intersection of equipotential surfaces

Question

I couldn't comprehend why equipotential surfaces cannot intersect. Now everywhere I searched, the reasons I got were either that the potential at the intersection would be different, or the field would have 2 direction. My doubt is,

1. If the fields have two directions, can't we simply add the vectors? Electric field is simply the force per unit charge, so if a charge is kept at that point, wouldn't it experience a force which is a resultant of both?

2. If the potentials are different at that point, why would they simply add up? Let's say both surfaces are at their own different potentials due to some external agent (say, a charge at the centre of a spherical shell). So when they intersect, shouldn't the net potential at the intersection be the sum of both the potentials?

Answer 1

Let's tackle your queries one by one.

1. If you proceed to find the resultant of the 2 electric fields (whose magnitudes are not known to you), the resulting direction will not be normal to either surfaces. As you know, the NET electric field at a point must be normal to any equipotential surface. Since that is unique at a given point, that is simply not possible.

Also note that, the presence of an electric field determines the equipotential surface, not the other way around. So associating an electric field with an equipotential surface doesn't make sense.

2. If you were to add the potentials, then the point of intersection would not belong to either surfaces, since the sum will not be equal to either potential.

Also not that while 2 different equipotential cannot intersect, the same surface CAN intersect itself (extend a lemniscate shape to 3D). The intersection point has the same potential as the rest, but the field at that point is zero, since this is the only electric field that can satisfy the condition of not having any tangential components to either surface.

Answer 2

Two different equipotential surfaces can't intersect because then they would have had the same potential and wouldn't be two different equipotential surfaces after all.

All points of an equipotential surface have the same potential, thus the name. So this issue is mainly about terminology. Sure, surfaces can intersect. But if they happen to be different equipotential surfaces, then they must have the same potential when intersecting - otherwise we wouldn't have called them different equipotential surfaces to begin with.

Compare this to simple equipotential lines on a map. Such lines correspond to level lines, height curves. All points are, by definition, at the same height on an equipotential line on a map. That's how hikers and orientation runners can figure out the 3D terrain from a 2D map. If two such equipotential lines were to intersect - meaning, if two paths each at constant heights were to intersect - then that means they must have the same height. Otherwise they would have missed each other, going above or under one another, not intersecting.

Answer 3

You can see this in two ways:

From the Potential: Consider that the electrostatic potential is a scalar field. It therefore necessarily has a single, well-defined, gradient at each point, which determines the direction of the electric field. It is therefore impossible for there to be two electric fields at a point. While you can decompose the electric field at the point into two vectors (or 20 vectors, or whatever), those are not the physical electric field, which is only one. So since the electric field is perpendicular to the equipotential line, the fact that there is a single field (gradient direction) implies there can be only one direction to the equipotential line at that point (the direction perpendicular to the gradient).

The only exception is if there is a region/point with no gradient. A level-plane surface. In this region the field will be zero, and talking about equipotential lines makes no sense.

From the field: There is one value for the electric force at a point, and therefore for the electric field at that point. Because the divergence of the electric field is zero (Maxwell's equations), this electric field must be the divergence of a scalar field, which we call the "electrostatic potential". And then continue as above.

Whichever you deem to "really exist" - the electric field, or the electrostatic potential - it follows that at every point the field is the gradient of the potential, and hence is perpendicular to its equipotential lines (the perpendicular direction to the gradient) and hence two such lines cannot cross. Except in a region where there is no gradient at all, and hence no field and no equipotential lines.