Ok, consider this sign convention for change in potential energy.

If $ \Delta U = + $ : There is increase in potential energy
If $ \Delta U = - $ : There is decrease in potential energy

Now, lets's move onto the main part of the post.

Consider 2 point positve charges, Q as a source charge and q as the test charge.
Now, the Q charge will have electric field, $\vec{E}$ going towards to $\infty$.
The datum is considered $\infty$ where $ U_{\infty} = 0 $.

The point charge, q is brought from $ \infty $ to a distance, r from Q charge. While bringing it from $ \infty $ to a distance, r, an external force, $ \vec{F} $ acts on it which is equal and opposite to the electric force $ \vec{F_e} $

So, $ \vec{F} = -\vec{F_e} $ and $ F_e = k \frac{Qq}{r^2} $

Let the work done by the electric force be $ \,dW_e $. So,

$$ \int dW_e = \int_{\infty}^r \vec{F_e}\,d{\vec{r}} $$ $$ W_e = q\int _{\infty}^r \vec{E}\, d{\vec{r}} $$ $$ W_e = -q\int_{\infty}^r E\,dr $$ $$ W_e = -kQq\int_{\infty}^r \frac{1}{r^2}\, dr $$ $$ W_e = -kQq\biggl[\frac{-1}{r}\biggr]_{\infty}^r $$ $$ W_e = k\frac{Qq}{r} $$

Now, as $ \vec{F_e} $ is a conservative force, so

$ W_e = -\Delta U $


$$ {\color{red} {\Delta U = -k\frac{Qq}{r}}} $$

Now, Q, q, r : All these variables are positive.

$ \Delta U = Negative$

Thus, this means that Final Potential Energy < Initial Potential Energy or to put in other ways potential is decreased when it is moved from $ \infty $ to distance ,r.

Now, this is supposed to be a contradictory result. Why??

Because one can easily see that $\vec{E} $ is going towards $ \infty $, the direction of decreasing potential/potential energy.
And because, $ U_\infty = 0 $ so, Ur should certainly be positive.

But this is not what the result is telling me. Why such contradiction is there in the result?

Using same sign conventions do not create any problems in case of negative source charges ,i.e, -Q.

So, please help in clearing this doubt. Also, please go through the next section.

Some Extra Points Related To This

-----> First of all to notice is that the angle between $ \vec{E} $ and $ \, d{\vec{r}} $ is 180°.

The angle is 180° because there is an external force, $ \vec{F} $ acting equal and opposite to the electric force, $ \vec{F_e} $ along which the charge, q moved(i.e., displacement is opposite to electric force).

The purpose of external force is to just balance the electric force so that there is no net increase in kinetic energy of the test charge. Otherwise, in the absence of external force, while evaluating the work done, $ W_e $, I had to consider the net kinetic energy also which would also change the expected value of potential energy at the distance, r.

-----> The most expected answer I think, I will get is that "you have set the parameter/sign conventions of potential energy like that". If you swap the signs, everything will get resolved.

Ok, see, if I even swap the signs of potential energy given at the very beginning of the post, it will resolve the problem/contradiction ocurring.

But, if I swap the signs what it will do that in case of negative source charges(i.e., -Q) same contradiction will occur but the signs will get reversed in the value of potential energy.

In that case, the external force will move the positive test charge, q from distance, r to $ \infty $ opposite to the direction of attractive electric force $ \vec{F_e{_A}} $. So the limits would be like $ \int_r^{\infty} \vec{F_e{_A}}\, d{\vec{r}} $

Lastly, I know there are many experts in this community. No disrespect, but please don't write that "the sign of change in potential energy don't matter at the end".

According to me the sign of change in potential energy matters as long as you are stuck with one set of sign conventions but the sign of potential energy(at a point) doesn't matters. If I am wrong at this point, please correct me.


In case of +Q charge, the external force will move the test charge,q from $\infty\rightarrow r$ whereas for -Q charge, it will take it from $r\rightarrow \infty$.

Due to the conservative nature, the electric force does 0 work in closed path. Thus,

$W_{\infty\rightarrow r}=- W_{r\rightarrow \infty}$

Considering Bill N statement that the differential element $d{\vec{r}}$ is always points from origin to $\infty$ then,

For +Q charges, the repulsive force $\vec{F_e}$ is parallel to $d{\vec{r}}$ and for -Q charges, the attractive force $\vec{F_e}$ is anti-parallel to $d{\vec{r}}$.

Now, as the direction of differential element is handled by limits of integration(Bill N comment) then, it's for sure that for either of the 2 cases(+Q and -Q source charges), the scalar product between $\vec{F_e}$ and $d{\vec{r}}$ should be same.

Earlier, in this post, for +Q charges, the work done by repulsive force was coming out to be $W_{\infty\rightarrow r} = k\frac{Qq}{r}$ and so the $\Delta U= -k\frac{Qq}{r}$

A counter-intuitive result if $\Delta U= U_{final}-U_{initial}$ While evaluating the $W_{\infty\rightarrow r}$ the negative sign in the result of integration was cancelled out by the negative sign of scalar product.

I have done the same for -Q charges and the expression was $W_{r\rightarrow \infty}= -k\frac{Qq}{r}$ and so the $\Delta U= k\frac{Qq}{r}$.

Again, an invalid result if $\Delta U= U_{final}-U{initial}$. Here, the negative sign of scalar product was the only negative sign.

I did all of this was done again but taking angle between the $\vec{F_e}$ and $d{\vec{r}}$ equal to for both types of source charges. This time, the results were opposite and signs of change in potential energy were absolutely correct(if $\Delta U= U_{final}-U_{initial}$).

So, the only thing which is striking my mind is why the angle between the 2 when taken 0° is giving right signs in $\Delta U$ even though for negative source charges, the $d{\vec{r}}$ is opposite to $\vec{F_e}$?

  • 4
    $\begingroup$ I don't think there's any expert who would say that the sign of the change in potential energy doesn't matter. If someone says that, they are not an expert, and don't deserve any particular respect. $\endgroup$ – Mike Feb 20 '18 at 17:37
  • $\begingroup$ What is the question, exactly? You say "this is supposed to be a contradictory result. Why??" Who supposes that? Are you saying you think it's contradictory? Are you asking why potential decreases when moving the charge $q$ from $\infty$ to $r$? Will you accept Wikipedia's statement that it is "tradition to define this function with a negative sign so that positive work [done on an object by the field] is a reduction in the potential"? $\endgroup$ – Mike Feb 20 '18 at 18:04
  • $\begingroup$ I have frequently struggled with this integration. I don't know why your analysis doesn't work, I suppose it has something to do with integrating in the "wrong" direction. There surely is a way to fix your approach. The way that I get this to work is by integrating from $-\infty$ to $-r$. Then my integration goes "from left to right", which is what I'm used to. $\endgroup$ – garyp Feb 20 '18 at 18:59
  • $\begingroup$ Is there anyone who can clear my confusion stated at the end of the question??? $\endgroup$ – lakhi Feb 23 '18 at 18:16

Your mistake is in the third integral equation and your consequent statement later:

-----> First of all to notice is that the angle between E⃗ and dr⃗ is 180°.

No, it's not! When you set up $\int \vec{E}\cdot \mathrm{d}\vec{r}$ you must keep in mind that the $\mathrm{d}\vec{r}$ must be consistent with the limits, too. The ordering of the limits, not the differential vector element, tells you what direction you are moving when you do the integral. Let's consider the following 2 integrals $$\int\limits_3^6 x ~\mathrm{d}x \hat{i} \text{ and} \int\limits_6^3 x~ \mathrm{d}x \hat{i}.$$

They must be the negative of each other. So, the ordering of the limits determines what the integral means. The differential element is necessarily interpreted as defining the positive change direction for the increment.

In other words, $\mathrm{d}\vec{r}$ points out from the origin toward $+\infty$ and is parallel to $\vec{E}$. That's where your incorrect negative sign comes from.

  • $\begingroup$ I am really not getting this point "The differential element is necessarily interpreted as defining the positive change direction for the increment". Can you please explain clearly this point? Are you saying that the differential vector element, $\, d{\vec{r}} $ always point towards electric field whether there are positive source charges, +Q or negative sources charges, -Q? $\endgroup$ – lakhi Feb 21 '18 at 7:11
  • 1
    $\begingroup$ In order to do any calculations with vectors one must define a coordinate system. The differential elements used in integration (dx, dy, dz, dr, d$\theta$, etc) are considered to be positive themselves, and if an integral is to be done which needs to have a variable changing in the negative direction, that negative change is handled by the limits of integration. See my example integrals in the answer. $\endgroup$ – Bill N Feb 21 '18 at 13:23
  • $\begingroup$ So $\mathrm{d}\vec{r}$ points from the origin to $\infty$ no matter what quantity you are integrating. $\endgroup$ – Bill N Feb 21 '18 at 13:25
  • $\begingroup$ I have added a small doubt related to this. Please have a look. $\endgroup$ – lakhi Feb 22 '18 at 20:08

I think you are confusing what does the work, i.e. if the work is being done by the electrostatic field or by the external agent.

If you have a positive test charge $+q$ and you (i.e. the external agent) bring it from infinity to a distance $r$ from a positive charge $+Q$, then the work done by you (the external agent) is

$$W_{ext}= \int_{ \infty }^{r} dW = \int_{ \infty }^{r} -F_{ext} dx= -kQq\int_{ \infty }^{r} \frac{1}{x^2} dx=-kQq \left [ -\frac{1}{x} \right ] _{\infty }^{r} = +\frac{kQq}{r}$$

If you are confused about the signs, you can just consider the one-dimensional problem (which I basically considered above), with $Q$ at the origin and $q$ being moved inwards from right to left. The electrostatic force on $q$ due to $Q$ points from left to right (i.e. it is repulsive) and therefore it has a positive sign. The external force must be the opposite of it: being directed from right to left, it is negative. However, the minus sign cancels out with the minus resulting from the integration.

Indeed, we could have expected the work done by the external agent to be positive, since work has to be put in in order to unite two like charges.

The electrostatic potential energy shared by two charges is the work required to bring two charges from an infinite separation to a separation of $r$. Indeed, this is the work we found above: therefore, the electric potential energy shared by two positive charges (and, more generally, two like charges) is always positive, except of course at infinity where it is defined to be zero.

I don't fully understand what your question is but I hope this does at least some good in answering it.

On another note, two "fussy" remarks on your notation. 1) The integrand of the expression for work in two or more dimensions features the dot product notation, which you omitted -- it makes no sense to just write two vectors next to each other, you must explicitly write what kind of product you are performing between them -- hence it should read $\underline F \cdot \underline {dr}$ rather than simply $\underline F \underline {dr}$, as this latter expression makes no sense. 2) Also, just to do justice to mathematics, avoid using the same variable as the integration variable and in the limits, i.e. it would be better to write $\int_{ \infty }^{r} \frac{1}{x^2} dx$ rather than $\int_{ \infty }^{r} \frac{1}{r^2} dx$. Sorry for being fussy, just thought you might find these remarks useful in the future.

  • $\begingroup$ Thanks for suggestions.Ok I think you choose a positive direction along the direction of electric force, i.e., $r\rightarrow\infty$.Also, your evaluation of work done by external force is also correct.It is positive.Now, suppose at the place of Q source charge you have -Q source charge. This time, the positive direction will be $\infty\rightarrow r$ because the electric force is directed like this. This time also the work by external force should be positive as the test charge would move into it's direction but this time the evaluation of integration $\int_r^{\infty}$ would give negative work. $\endgroup$ – lakhi Feb 20 '18 at 19:40
  • $\begingroup$ Try making this argument for a system of a positive and a negative charge initially $1.0$ m apart and released from rest. There is no external force pulling them together, but the potential energy still changes. The change in potential energy is the negative of the work done by the conservative force, not by any external force! $\endgroup$ – Bill N Feb 20 '18 at 20:15

Ok, I think I got it but it may become hard for me to explain.
Things gets started to wrong when I assumed that Final Potential Energy < Initial Potential Energy.

It means that $\Delta U = U_f - U_i $.
But if I take $\Delta U = U_i - U_f $ then everything goes right with the situation, i.e., Final Potential Energy > Initial Potential Energy and thus, no contradictions.

See, the point is that the sign of $\Delta U $ you will get entirely depend on the limits of integration.
Swap the limits and the sign of $\Delta U $ will get reversed. For more information relating to the swapping of limits of integration refer to this page enter link description here

So, the sign in the change of potential energy, $\Delta U $ doesn't explicitly concludes that there is increase in the potential energy or decrease in potential energy(which I what I did in my question).

The conclusion that there will be increase or decrease of potential energy between any 2 points depends how you take the difference between the potential energy of 2 points(it could be $ U_f-U_i $ or $ U_i-U_f $).

Also, while finding the potential energy at a point, either one of them($U_f$ or $U_i$) may be the datum/reference point.

  • $\begingroup$ No, you can't change the definition of $\Delta U$. It is most definitely $U_{after}-U_{before}$. All delta quantities are defined like that, and the sign is important. $\endgroup$ – Bill N Feb 20 '18 at 20:11

protected by Qmechanic Feb 20 '18 at 20:17

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?

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