# Why there is a negative sign in the formula of calculating work done in electrostatics?

$$W = - \int _ { a } ^ { b } \vec { F } \cdot d \vec { r }$$
( from The Feynman lectures on physics vol. 2, new millennium edition, page 4-4)

In this formula why there is a negative sign in the formula? I am not asking the sign of the total workdone at the end of the calculation (which is already answered in this question which is similar to mine) but the sign in the formula that is before calculation. from where the sign comes from?does it have anything to do with the co-ordinate system,if yes then can the sign be changed if we take a different quadrant?. • How is $\vec{F}$ defined in this problem? – The Photon May 24 at 16:00
• The $\vec { F}$ here is the applied force – Hawkingo May 24 at 16:01
• Then if I change the coordinate, for example, the other charge is moving in q3 quadrant ( in 2 dimensions, (x and y are negative)) then the dr will be positive, so how can a formula change /depends on a coordinate? – Hawkingo May 24 at 16:09
• it isn't the applied force it is the "electrical force". Please see my answer. – The Photon May 24 at 16:33

The key is this:

The work done against the electrical forces in carrying a charge along some path is the negative of the component of the electrical force in the direction of motion. [emphasis added]

Generally, if we apply a force $$\vec{G}$$ to some object and move it along a path, the work done on the object is $$W = \int_a^b\vec{G}\cdot d\vec{s}$$

In this case, the force we must apply to the charged particle to move it quasistatically is opposite to the force $$\vec{F}$$ being applied to it by the electric field,

$$\vec{G}=-\vec{F}$$

so

$$W = -\int_a^b\vec{F}\cdot d\vec{s}.$$

To restate it, in the description in Feynman, $$\vec{F}$$ isn't the force we apply to the particle, it's the force applied by the electric field that we have to work against.

• If that is the case then the work done by the applied force when bringing a positive charge from infinity to a certain distance (for example) would be negative (because if the negative sign is absent while calculating the work done by the applied force, then there would be a extra negative sign due to the integration of dr over the path). – Hawkingo May 24 at 16:59
• @Hawkingo Bringing a positive charge from infinity to a certain distance from what? A positive charge? Then, the work done by the external force would be positive. I am not sure what you mean by an "extra" minus sign "due to the integration of $d\vec{r}$ over the path". – Dvij Mankad May 24 at 17:37
• @Hawkingo, If you're bringing a positive test charge from vicinity towards a positive charge at the origin, then $\vec{F}$ (the force due to the electric field) points away from the origin, in the opposite direction of your motion. So $\vec{F}\cdot d\vec{s}$ is negative, and you need the negative sign in front of the integral to get the expected result of positive work done on the test charge. – The Photon May 24 at 17:39
• @Hawkingo, I'm not clear why you changed $d\vec{s}$ to $d\vec{r}$ when you transcribed the equation from the text into your question. The usual interpretation of $\vec{r}$ is not the same as $\vec{s}$. $vec{r}$ is the polar coordinate vector away from the origin. $d\vec{s}$ is an element of the path you're pushing the particle along. If you're bringing the particle from infinity, these two vectors point in opposite directions. – The Photon May 24 at 17:42
• @Hawkingo It looks like you might be getting the sign of the differential confused. See my answer here that deals with that – Aaron Stevens May 24 at 23:26

The force on a particle can also be described in terms of a scalar potential function $$U(x)$$, it has nothing to do with coordinate frames.

\begin{align} F(x)=-\frac{dU}{dx};\hspace{5mm} \vec{F}(\vec{r})=-\vec{\nabla}U(\vec{r}) \end{align}

The potential energy $$U$$ means the amount of work needed to move an object from some point to another, the force applied needs to be equal but opposite, thats why there is a negative sign.

The force exerted by the force field always tends toward lower energy and will act to reduce the potential energy.

The negative sign on the derivative shows that if the potential U increases with increasing x, the force will tend to move it toward smaller x to decrease the potential energy.

Well not only in electrostatics but in whole classical mechanics "The change in potential energy of the system is defined as the negative of work done by internal conservative forces of the system".

• Yes. Definitions of (change of) potential energy in terms of work done by an external force lead to problems, confusion (such as that generated by Feynman), the need to invoke Newton's third law, the need to add some wording about how the force is applied (constant v? quasistatic?) , the need to specify that no other forces are acting, and the need to consider phenomena outside of the system in question (the external force). BTW, there are a lot of things in Feynman that need to be ignored or glossed over. Unfortunately, you need to know the subject in advance to know which things. – garyp May 24 at 16:47
• @garyp I know the subject, in fact the same thing is written in also different physics books for example electricity and Magnetism by E. Purcell. – Hawkingo May 24 at 17:01
• @unique As I said in my question, I am not talking about the sign of the work done but the sign in the equation which is eventually cancelled out by another negative sign originated by the integration of the dr over the path(when charge is moving towards the field/test charge) – Hawkingo May 24 at 17:03

My guess is - Work done is being done in the opposite direction of force which is created by electric field(electrical repulsion) . The definition which is interpreted from my book is that work done in bringing a test charge from infinity to that point without producing any acceleration. So the net force is being applied in the opposite direction of displacement.you can imagine you bringing a charge from infinity but without any acceleration

OK ,I got it.but doesn't potential decreases as the distance from the source/field increases and vice-versa?so if the charge is travelling from a towards b it's potential should increase,right?please clear my misconception

the minus sign will be resolved after doing the integral for ex - for a point charge net work done = ∫dw = ∫-kq/r^2 dx = kq/r

I agree with @The Photon answer.

First, as @The Photon noted, the force $$F$$ in the formula is the force that the electric field exerts on the charge. Consequently the negative sign in the formula is for the work done by an electric field where the force of the field on the charge is opposite to the direction of movement of the charge. This means the electric field is doing negative work on the charge taking energy from the charge and storing it as electrostatic potential energy (increasing its electrical potential). The charge gets that energy from an external electrostatic force that does an equal amount of positive work on the charge. The gravity analogy is taking a mass at rest off the ground and placing it at rest a height $$h$$ above the ground. You do positive work $$mgh$$ to move the mass. Gravity does an equal amount of negative work $$-mgh$$ taking the energy you supplied and storing it as gravitational potential energy

Note, however, that if instead the electrostatic force in the same direction as the movement of the charge, the field would be doing positive work decreasing the electrostatic potential energy of the charge and giving it kinetic energy. The gravity analogy is the positive work done by gravity on a falling object.

So to a certain extent the selection of the sign is arbitrary. However in electrostatics the potential at an infinite distance from a fixed positive point charge is arbitrarily assigned zero potential. This is logical since the strength of the field of the point charge approaches zero an infinite distance away. Now if another “test” charge at that infinite distance is to be moved toward the fixed positive charge its potential will either increase or decrease depending on the polarity of the test charge. If the test charge is negative, in moving it towards the fixed positive charge the field will do positive work and the potential of the charge will decrease. If the test charge is positive, the field will do negative work and increase its potential. The latter is the basis of the formula.

Hope this along with the other answers helps.