0
$\begingroup$

Before you tell me there are a bunch of other similar questions asked as well, hear me out ;(

I'm really confused when it comes to work being done. When you're considering a charge moving from a one point to another against an electric field, you consider TWO types of work being done. One is the work being done by the external force (which is always positive, since cos0=1) and the other is the work being done by the electric field (which is always negative, since cos180=-1).

What I don't seem to understand is, whilst deriving the formula of potential gradient, why do we seem to mix up the work due to the electric field and work due to the external force? If we were to only consider work being done by the external force, technically, the formula for the potential gradient should be positive. But, in ALL the books, they seem to add in a negative sign by simply stating, "The negative sign indicates that the work being done is against the field force."

That should technically mean that they're considering work done due to the applied force, and according to the definition of work, it should be positive! But if you say just equate the external force to the electric force (since the external force is the negative of the electric force), then you'd also have to change cos0 to cos180 and thus, the negative sign would vanish again.

Which work do we consider? The one due to the external force or the one due to the electric field? Why? Is the negative sign just a convention to show us that we're actually considering the work due to the external force and that force acts in a direction opposite to the field force?

TL;DR: What is the mathematical explanation behind the origin of the negative sign (in terms of the cosine of the angles)? Where did it come from?

UPDATE: Philip's and Overwootch's answers give a pretty straight-forward mathematical explanation of where the negative sign popped up from, in case you were to traverse the charge in the direction of the electric field, or want to consider the work done by either the electric field or the electric force.

Thanks, everyone!

$\endgroup$
4
  • $\begingroup$ the minus sign is NOT by convention, but by DEFINITION. ΔU=-work done by conservative force. This is how potential energy for a conservative force is defined. Differentiating both sides automatically shows you that (conservative) force is the negative derivative of potential energy $\endgroup$ – OVERWOOTCH Feb 19 at 14:38
  • $\begingroup$ I don't know why this relation is always introduced in electrostatics. This relation is true for any conservative force in general (gravitational, electric, nuclear, spring....) $\endgroup$ – OVERWOOTCH Feb 19 at 14:39
  • 1
    $\begingroup$ Definitions are conventions $\endgroup$ – Dale Feb 19 at 16:27
  • $\begingroup$ @OVERWOOTCH That's completely fine, I just don't understand HOW the MATH works in bringing about this negative sign when you're considering work done BY the external force. $\endgroup$ – ihateelectricalphysics Feb 19 at 17:08
1
$\begingroup$

I think I understand your confusion. You can define the potential difference between 2 points in 2 ways.

  1. The electric p.d between 2 points in an electric field is the negative of the work done by the electric force in bring a unit positive charge from the initial to the final point. This follows from how potential is actually defined mathematically in general
  2. The electric p.d between 2 points in an electric field is the work done by an external force AGAINST the electric force in bringing a unit positive from the initial to the final position WITHOUT CHANGING ITS K.E. Now note the phrases in italics. "Against" &"without a change in k.e" imply that the external force is equal in magnitude and opposite in direction to the electric force. I.e Fext=-(unit charge)(E)[since Fext & E are vectors, the minus sign indicates the opposite direction]

so ΔV=FextΔX=(-E)ΔX

The important thing to keep in mind is that the potential difference is actually defined interms of the work done by the electric force NOT that done by some external force. Infact, there is no external force. The proper method of calculating ΔV is to release a unit positive charge from the initial to the final position, calculate the work done by the electric force (through gain in k.e maybe), and put a negative sign on it.

However, in most textbooks, explaining the addition of the negative sign in the definition above is difficult so they try to rephrase the definition to get rid of it. They introduce a hypothetical "external force" that is equal in magnitude and opposite in direction to the electric force. The unit positive charge is then moved again from the initial to the final position, this time under the influence of the electric and the hypothetical force. Since the hypothetical force is equal in magnitude and opposite in direction, it's work done is already equal to negative of the work done by the electric force, so you just put that in the definition.

Both definitions are equivalent, but the second one is just a handwavy way of making the original one more intuitive through the introduction of a hypothetical external force

$\endgroup$
5
  • $\begingroup$ However, here's where I'm confused. Let's say you're calculating the work done by the external force. So, the formula would go something like Fr (r being the displacement being traversed). HOWEVER, if you were to replace the applied force by the electric force, it would be something like -Fr. And since work is a dot product, you'd use cos180 to get another - sign which would then result in a positive work done. How will you add a negative sign in that case? The - sign seems perfectly placed in case of work done by the field force or if the charge were to travel in the direction of the field. $\endgroup$ – ihateelectricalphysics Feb 19 at 20:35
  • $\begingroup$ in all of the answers here, ΔX ιs assumed to be in or against the direction of the electric field and thus a positive/negative scalar. None of the answers use the dot product definition here. $\endgroup$ – OVERWOOTCH Feb 19 at 21:16
  • $\begingroup$ That was one of the things I was confused about the most haha, but I've pretty much understood most of it. Thanks! $\endgroup$ – ihateelectricalphysics Feb 19 at 21:19
  • $\begingroup$ could you post a picture of the derivation of the potential gradient. I think I finally understand the issue $\endgroup$ – OVERWOOTCH Feb 19 at 21:26
  • $\begingroup$ Sure, here you go: imgur.com/58p7cGa Might be a bit confusing because it's pretty weird. $\endgroup$ – ihateelectricalphysics Feb 19 at 21:50
4
$\begingroup$

If you don't have any external force, use the work done by electric field (to be precise is the work done by the electric force $F = qE$), if instead you have a external force, conservation of energy cannot by applied (an external work is been done), here is an example.

You want to decrease the distance between two positive charges $q_1$ and $q_2$. For doing so you need the work of an external force $F_{ext}$. If you wouldn't have $F_{ext}$, one of the charges particle moving with speed $v_o$ would see a decrease in its kinetic energy and an increase of it's potential energy (this is possible by the work done by the electric force, which can "tranform" kinetic energy in potential energy and viceversa being conservative). If you have an external force that equals the electric force and the distance get smaller, the speed $v_o$ remain the same but the potential energy is increased (condervation of energy is not valid due to the external work been done). So you need to consider both work/forces($F_{ext}$ for not being conservative, and the electric force due to its potential).

the minus in $E = - \nabla V$ is only a convention, infact the definition is

\begin{equation} -\Delta V = V_{a} - V_b = \int_{a}^{b} E \,ds \end{equation}

$\endgroup$
9
  • $\begingroup$ So, there's no way to explain the sign using the cosines of the angles? $\endgroup$ – ihateelectricalphysics Feb 19 at 12:53
  • $\begingroup$ yeah not a deep matematical explanation. worth saying that that minus is "comfortable" when you write U as qV $\endgroup$ – lorenzo Baldessarini Feb 19 at 12:57
  • $\begingroup$ using sin and cos can lead to some confusion because the angles aren't always the same (depend by the charges of the particles). in this case external force and movement have the same direction, so the work done is positive, and the eletric force do a negative work(eletric force point "outwards" but the movement is "inwards"), infact the kinetic energy decreasce (work can be seen as variation of kinetic energy) $\endgroup$ – lorenzo Baldessarini Feb 19 at 12:58
  • 2
    $\begingroup$ I agree with Iorenzo. It is just a convention. It could have been defined the other way, and all that would change is that we would flip the sign of any formula with $V$ $\endgroup$ – Dale Feb 19 at 13:03
  • 1
    $\begingroup$ @J.Murray True I guess. I just wanted to point out that this choice is not as arbitrary as choosing which charge to call positive, or defining electric field to be the force on a unit positive charge as opposed to a unit negative charge $\endgroup$ – OVERWOOTCH Feb 19 at 18:19
1
$\begingroup$

I don't know why so many textbooks bring external forces into their treatment of electric potential. Here's a simpler treatment...

Work done by field on test charge $q$ going a small distance $\Delta x$ is $qE_x\ \Delta x$. But this is the loss in electrical potential energy, $E_p$, of $q$, as this amount of work is no longer available to be done. So $$\Delta E_p=-qE_x\ \Delta x.$$ Dividing through by $q$, remembering that, by definition, $\Delta V=\Delta E_p/q$, $$\Delta V=-E_x\ \Delta x.$$ The electrical potential energy lost by $q$ may become the kinetic energy of the particle carrying $q$, if no other forces are present, or it may be stored in some other form if work is done against some external force. But what exactly happens to the energy doesn't affect the derivation above.

$\endgroup$
10
  • $\begingroup$ You mean a CHANGE in electric potential energy, right? How is it decreasing? If we were to bring the charge closer to the positive plate and then removed the external force, the charge would repel away with an increasing kinetic energy coming from the stored potential energy, right? $\endgroup$ – ihateelectricalphysics Feb 19 at 13:20
  • $\begingroup$ $E_x$ is the component of electric field in the $x$ direction. So if $E_x \Delta x$ is positive work is done by the field on the charge, and the change in electrical potential energy is negative. $\endgroup$ – Philip Wood Feb 19 at 13:26
  • $\begingroup$ Okay, yeah, that makes sense. But, what if we were considering the charge to be moving in a direction opposite to the electric field (thus bringing in the concept of an external force)? $\endgroup$ – ihateelectricalphysics Feb 19 at 13:31
  • $\begingroup$ Exactly the same argument applies. But since $E_x \Delta x$ is negative, $\Delta V$ will be positive. I agree that an external force will be needed to make the charge go against the field, but that external force isn't needed in the calculation. $\endgroup$ – Philip Wood Feb 19 at 14:21
  • $\begingroup$ So, we're only considering the work done due to the electric field and not due to the external force? I mean, if we do it like this, the calculations make perfect sense. But, why can't we consider the work done due to the external force? Because I'm pretty sure in that case, you'd just have to stick to the idea of the negative sign being a part of the convention. $\endgroup$ – ihateelectricalphysics Feb 19 at 14:23
1
$\begingroup$

It is not only a convention, or at least, not an arbitrary one: it would be pretty odd to choose the sign the other way.

What is a potential? It's a function that tells you the potential energy per unit something (like mass or charge) of a widget when it's at a particular point in space. The source of the potential energy is a field, which is the force per unit something that acts on the widget. Where the widget has more potential energy, the potential is higher; where less, lower. Getting rid of the negative sign would mean that the force from the field pushes the widget toward higher potential: higher potential means lower widget potential energy. Why would you do that? You'd at least have to call it something other than "potential". "Notential" might fit the bill.

$\endgroup$
1
  • $\begingroup$ If the widget is being pushed to the higher potential, wouldn't it start storing higher electric potential energy? (also, I sort of now understand that the equation wouldn't work without the - sign, but I just don't understand the math behind how that - sign would appear when you're considering the work being done by the external force.) $\endgroup$ – ihateelectricalphysics Feb 19 at 17:13
1
$\begingroup$

The gradient of potential is defined to be the direction of increasing potential energy for a positive test charge. However, to increase the potential energy the external agent would have to move the test charge opposite the electric field. Thus the field is the opposite of the gradient.

$\endgroup$

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.