Asked by: Joe

Let us first state what these quantities, i.e. electric potential and electric field, mean.

The electric potential is a scalar quantity which quantifies the amount of electrical potential energy a unit test charge would have if it were placed at a specific point in space. ('Test charge' means that the charge is brought to that location without disturbing the other charges.)

In contrast, the electric field is a vector quantity (meaning it has both a magnitude and a direction). It quantifies the force a unit test charge would experience if it were placed at a specific point in space.

These two quantities are related very closely, since force is the negative of the derivative of the potential energy, the electric field is the negative gradient (gradient, simply put, is a fancy name for derivative in three dimensions) of the electric potential.

In other words, the electric field is a measure of how quickly the electric potential changes, and it points in the direction of maximum negative change of the electric potential. Thus, the value of the electric potential at that point is not directly related to the electric potential at that point. What matters is how fast the electric potential _changes_ around the point in question.

In the example given in the question, the potential is zero in the mid-point, but it changes quite rapidly when one moves towards one charge or the other. So, although the electric potential is zero, the field is nonzero.

Let me try to explain by analogy. The case of a car on a road is analogous to this. The elevation of the car from sea level determines is analogous to the electric potential (it is the gravitational potential), and the force on the car is analogous to the electric field. Now, the force on the car is determined by the slope of the road (derivative, gradient) and not by the elevation of the road from sea level. In other words, if your car was on a road which is sloping down, it _would_ experience a downhill force even if at that moment it happened to be at exactly sea level (i.e. the potential was zero). In fact, the force depends only on the slope, and it would be unaltered if the same road was carried to a different elevation. (Well, almost, since gravity does change with altitude, but I hope you get the point.)

There is another point... The 'zero point' of any potential is arbitrary. (At least I do not know of any counterexample.) Using a certain reference may be _convenient_, (such as no charges around = zero potential or sea level = zero potential) but any other choice works just as well. It may make the math more complicated, but it will produce the same physical results. This is because only _changes_ in potential have any physical significance.

Answered by: Yasar Safkan, Ph.D., Software Engineer, Noktalar A.S., Istanbul, Turkey

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'Physics is mathematical not because we know so much about the physical world, but because we know so little; it is only its mathematical properties that we can discover.'**Bertrand Russell**

(*1872-1970*)