Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

I can't conceptually visualize why it would be so. Say you have two point charges of equal charge and a point right in the middle of them. The potential of that charge, mathematically, is proportional to the sum of their charges over distance from the point ($q/r$). But intuitively, my thought process keeps going back to the concept of direction and how the electric field at that point would be zero. So why would the electric fields cancel while the electric potentials just add up algebraically?

share|improve this question
    
I do not yet fully understand what you actually want to know. Electric field being zero, is related to a minimum in the electric potential. –  Bernhard Feb 23 '13 at 23:22
add comment

3 Answers

Let me first comment that the statement

electric fields cancel while the electric potentials just add up algebraically

is not actually correct. Electric fields add due to the principle of superposition (see the section on superposition in the wikipedia article).

However, when two electric field vectors are of the same magnitude but point in opposite directions, then their sum is zero; this is what is happening at the midpoint between two equally charged particles.

Given an electric field $\mathbf E$, the electric potential $\Phi$ is defined through the relation $$ \mathbf E = -\nabla \Phi $$ so it is a scalar by definition. The electric potential also obeys the superposition principle. Provided we set the zero of potential at infinity, the potential due to a point charge $q$ is given by $q/(4\pi\epsilon_0 r)$, and $r>0$, so the potential of a point charge is either everywhere positive or everywhere negative depending on the sign of the charge. Therefore, given two point charges of the same sign, the sum of their potentials will cancel nowhere.

share|improve this answer
    
Expanding on this, the electric field is, mathematically speaking, the gradient of the electric potential. So an electric field of 0 does not mean an electrical potential of 0, it simply means that at that particular point in space, the electrical potential is not changing. –  Ataraxia Feb 24 '13 at 6:37
add comment

From a more general point of view, electric potential is not a scalar, but a component of a 4-vector (http://en.wikipedia.org/wiki/Electromagnetic_four-potential ) - it is not invariant with respect to boosts.

share|improve this answer
    
Well the potential is just one component of the four-vector. Not the four-vector itself. So isn't it a scalar after all? –  fffred Aug 23 '13 at 21:51
    
@fffred: No, the way physicists really think about this is that categories like vector and scalar are defined in terms of their transformation properties. A scalar is something that doesn't change at all under any smooth change of coordinates, e.g., a Lorentz boost. –  Ben Crowell Aug 23 '13 at 22:31
    
@Ben Crowell : I guess one can say that electric potential is a scalar under the rotation group and a component of a 4-vector under the Lorentz group. –  akhmeteli Aug 24 '13 at 10:37
add comment

An explanation based on the definition of scalar quantities in physics.

To see why electric potential energy is a scalar quantity you need to understand the following:

A physical quantity is a scalar property of a system, when its value and its effects do not depend on the orientation of the system.

Let us assume we have a system of three electrically charged particles carrying electric charge $+Q_A$, $-Q_B$ and $+Q_C$. Let us also assume the three particles are at positions $A$, $B$ and $C$ with position vectors $\bf {r}_A$, $ \bf {r}_B $ and $\bf {r}_C$ with respect to some arbitrary origin O. We can write the total potential energy of the system of three charged particles as

$E=\frac{1}{4\pi\epsilon_0}(\frac{ Q_AQ_C}{| \bf{r}_A-\bf {r}_C |}-\frac{ Q_AQ_B}{|\bf {r}_A-\bf {r}_B |}- \frac{Q_BQ_C}{| \bf {r}_C-\bf {r}_B|})$.

We can observe that as long as $|\bf{r}_{\mu}-\bf{r}_{\nu}|$ remain fixed, with $\nu\ne\mu$, the three particles can be placed in an infinitely large number of positions in various orientations, and yet $E$ will have the same value. I.e.

The orientation of the system does not bear any measurable effects on the value $E$.

This is why electrical potential/energy is a scalar quantity.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.