In case of electric potential at a point (P) the derivation says that it is equal to the external work done to bring a unit positive charge from infinity to point P without acceleration i.e with almost zero speed..the key word here is external..i.e a Force which would balance the force applied by field of source charge Q (placed at the origin) in direction opposite to vector R (in case Q is positive)..we can solve this by assuming that $dW = -F.dr$ and then go on to integrate and integrate and stuff..:)

question 1: what is the justification of the minus here..can someone clarify..if the coulomb law was not an inverse square law and say varied ∝ (r) i.e the force decreased as $r$ decreased..would the same minus sign apply here..Or is the sign simply minus as direction of external applied force is opposite to direction of increasing (r).

Question 2: Another confusing thing is that..let us say i displace the unit charge by a distance $dx$ when it is at any distance..since i am applying external force in direction of the source charge $Q$, the small amount of work done(by the classic $F.d$ formula) by external force is positive as angle between dx and F is zero radians..but if we use $dW =-F.dr$ (as the derivation demands) then the small work done is negative.. how do these two work done values reconcile.. their signs are opposite.. is it because work done is frame dependent

Question 3: In case of $dW = -F.dr$..the small amount of work done is negative..now if we sum it manually (taking a lot of lot of time haha) for all points from infinity to point r..then the total should be negative too right..as $dW$ will be negative for all small shifts until unit positive charge reaches point P..but if we integrate the same expression i.e $dW = -F.dr$, we get a positive work done (hence positive potential for positive charge Q at origin) for displacing charge from ∞ to distance..

Now for scalar stuff like work done..integration can be thought of like a super quick and polished summation technique ..but how do the signs differ then..when using summation(theoretically) and actually integrating the expression?

  • $\begingroup$ Are you following a textbook. Which? Which page? $\endgroup$
    – Qmechanic
    Apr 6, 2020 at 8:28

2 Answers 2


There are charges that create the potential, which exert force $\mathbf{F}$ on the test charge and perform work $W$ when (slowly) moving this charge to infinity. And there are also external forces which do the opposite opposite, i.e. exert the opposite force $-\mathbf{F}$ and do the opposite work, $-W$. Distinguishing these two clarifies most of the questions, whereas a particular sign is a matter of convention - which of these forces/works is taken positive.

  • $\begingroup$ I understand that..but you did not answer the question I asked.. $\endgroup$ Apr 6, 2020 at 7:56
  • $\begingroup$ Think it over a bit: sign has to do with the force that is doing the work: the one created by the charges or the one that counteracts it. It has nothing to do with how potential depends on $r$. $\endgroup$
    – Roger V.
    Apr 6, 2020 at 8:02

The general relation between potential and a conservative force is $$\mathbf F(x) = - \nabla V(x) $$ Integrating along a path C, $$ V = -\int_C \mathbf F.d\mathbf s $$ The sign convention is natural as the direction of the force is opposite to the gradient of potential (c.f. gravity). This is the force applied by the field. If you think instead of the work done against that force there will be a change of sign. The signs in the equations are not changed by a different $V$, because that is taken into account in the gradient.

The answer to your questions seems to be entirely to think carefully about whether you are talking about the force applied by the field, or the work done by a force acting against that field.


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