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I am reading Berkeley Physics Course, Volume 2 (Electricity and Magnetism by Edward M. Purcell). I am in chapter $3$, page $92$, and the book discusses conductors.

The following is from the book:

Because the surface of a conductor [in Fig $3.2$] is necessarily a surface of constant potential, the electric field, which is $-\nabla \varphi$, must be perpendicular to the surface at every point on the surface

I have omitted the picture because it is not relevant.

What is the reasoning?

I understand that the potential, $\varphi$, is a continuous function, and since $E=0$ inside the conductor and since $E=-\nabla\varphi$ I get that $\varphi=0$ inside and on the surface (from continuity) of the conductor.

However, I don't understand the reason the book gives for explaining why the field is perpendicular to every point on the surface.

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    $\begingroup$ Note that this is only valid if the conductor is in equilibrium. (This chapter of Purcell is only referring to electrostatics.) $\endgroup$
    – user4552
    Aug 17, 2013 at 18:56

6 Answers 6

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Loosely speaking, the gradient of a scalar field (such as the electrostatic potential) points in the direction of that field's greatest change. Since no change occurs in the field when you go along the surface, the gradient shouldn't have a component in that direction.

Here is another intuitive explanation: Imagine for a moment that the electric field was not perpendicular to the surface. That means it has a component along the surface. Now, electric fields exert a force on charges, so now we have a force on the charges in the conductor along the surface of the conductor. This force isn't balanced by anything else, so it will then move the charge around. But that means that our system wasn't yet in equilibrium, since charge was moving around. In equilibrium, the charges must be at rest, and that can only be the case when there is no electric force along the surface, i.e., when it's perpendicular to it.

Note: You say that $\varphi = 0$ inside and on the surface of the conductor. That is not true. $\varphi$ is constant inside and on the surface of the conductor.

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  • $\begingroup$ Loosely speaking, the gradient of a field points in the direction of that field's greatest change. By field do you mean a scalar field? The way you've expressed this may be confusing to people, since normally in E&M we reserve the term field for the E and B fields. The following sentence doesn't explain that "the field" refers to (I assume) the potential...? $\endgroup$
    – user4552
    Aug 17, 2013 at 18:54
  • $\begingroup$ Good point. I've added a little disclaimer $\endgroup$
    – Lagerbaer
    Aug 17, 2013 at 22:35
  • $\begingroup$ We can apply the contradiction only if the conductor is bound to reach electrostatic equillibrium,do every conductor,left for some time,come to electrostatic equillibrium,if yes,why?I can think of an example(though it may be vague) that suppose electric field is not perpendicular at surface,hence it must have some component along surface which will apply force on charge until it reaches equillibrium,what if charge continues to move on surface for infinite amount of time? $\endgroup$
    – Dheeraj Gujrathi
    Dec 15, 2022 at 7:26
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Assuming that that electric field is not perpendicular to the surface, then there must be a component of the electric field that is parallel to the surface. Since the electric field is defined to the be gradient of the potential, the surface of the conductor would not have a constant potential.

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If it wouldn't be its tangential component would exert a force on the charges and they would move. The condition would then not be static. After some movement and redistribution of charges (when there would be no force on charges) the condition will again become static thus making the field only normal to the surface

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If the electric field lines were not perfectly normal, then there would be some tangential component, which would accelerate the charges in the conductor and rearrange the charge distribution. For there to be a uniform distribution, this tangential component must be zero.

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    $\begingroup$ Do you mean the charge distribution must be uniform? That's not true. Perhaps slight better would be to say that the charges would rearrange themselves until there is no tangential acceleration, i.e., no tangential field. $\endgroup$
    – garyp
    Jun 1, 2016 at 16:27
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electric field lines run from higher potential to lower potential(that's why for positive charge it is radially outward and for negative charge it is radially inward).Potential V=Er where E is electric field and r is the distance from charge.For given E, if we find locus of all point (i.e. at constant r), we get a surface that as at a constant potential. Meaning, if you get any charge from infinity to any point on this surface, equal amount of work is done. Thus if you find this locus it is always perpendicular because of the fact that electric field lines actually represent the gradient in potential.

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As the potential is equal, there will be equal amount of work done by the electrons to move from infinity to any point on the surface. If the surface has different potential, then the electron will be accelerated, then it will automatically become a non equipotential surface. Thats why it should be normal.

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