9
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Bernhard Feb 23 '13 at 23:22
9
$\begingroup$

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.

$\endgroup$
  • 2
    $\begingroup$ 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. $\endgroup$ – Ataraxia Feb 24 '13 at 6:37
5
$\begingroup$

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.

Edit: in answer to question:

Well the potential is just one component of the four-vector. Not the four-vector itself. So isn't it a scalar after all?

As user Ben Crowell says:

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.

but also we could say that electric potential is a scalar under the rotation group and more generally under the Galilean Group but only a component of a 4-vector under the Lorentz group.

$\endgroup$
  • 2
    $\begingroup$ @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. $\endgroup$ – akhmeteli Aug 24 '13 at 10:37
  • $\begingroup$ +1 for enlightening me, but this is just going to confuse the asker. :P $\endgroup$ – Blackbody Blacklight May 1 '14 at 10:53
  • $\begingroup$ @akhmeteli I feel like I want to add Ben and your comments to the answer: Ben is of course correct, but your comment explains why the high-school explanation is not altogether wrong. Both are excellent comments deserving to be part of the answer. Are you OK if I do that? (or maybe you want to do it) $\endgroup$ – WetSavannaAnimal May 1 '14 at 12:22
  • $\begingroup$ @WetSavanna: I don't mind if you edit my answer or write yours. $\endgroup$ – akhmeteli May 2 '14 at 0:19
  • $\begingroup$ @Blackbody: On the one hand, answers are not just for askers, on the other hand, what can I do if simple answers are just not quite adequate?:-) $\endgroup$ – akhmeteli May 2 '14 at 0:23
3
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

Because work and charge both are scalar quantity ie electric potential is scalar quantity

$\endgroup$
0
$\begingroup$

When we bring a test charge let us say (+q) to a certain point, we exert some force. Our exerted force has to be equal or greater to the force exerted by the electric field in order to overcome it. Thus the forces being opposite and equal cancel each other and hence the direction is undetermined. Since it is a compulsion for vector to have both magnitude and direction, hence here there is no direction but magnitude and hence we call it a scalar quantity.

$\endgroup$
  • $\begingroup$ the question is not answered here... $\endgroup$ – Bruce Lee Feb 10 '16 at 9:22
0
$\begingroup$

Suppose in a equilibrium system where two negative charges is accompanied by one positive test charge in the middle. Here the resultant electric field at the centre is zero. Due to opposite and equal electric line of force from two negative charges but to bring that test charge from zero potential to the middle some work is done which is nothing but the potential energy value you get by adding algebraically. Though I am commenting in 2016 feel free to give me positive rating. :)

$\endgroup$
  • $\begingroup$ Welcome on Physics SE :) I don't think that asking for up-votes will be insanely well received here - just as a friendly suggestion ;) $\endgroup$ – Sanya Nov 28 '16 at 18:51
  • $\begingroup$ @Sanya maybe you are right. I can see that I am the only one who asked for it.Thanks.. $\endgroup$ – Abhishek Dec 16 '16 at 16:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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