# Sign of work against an electrostatic field

I think I'm slightly confused with the signs. The work needed in order to bring point charges $$q_i$$ from infinity to a distance $$r$$ from some shape of charge $$Q(r)$$ with a varying charge (because we bring more and more charges to it) is:

$$dW=k\frac{Q(r)dq_i}{r}\implies W=\int_{q_i} k\frac{Q(r)dq_i}{r}$$

The formula $$W=k\frac{q_iq_j}{r_{ij}}$$ was derived as the work done in order to bring a point charge $$q_i$$ to a distnace $$r_{ij}$$ from a point charge $$q_j$$ ,from infinity.

We will get some $$dr$$ in the integral, because $$dq_i=\rho4\pi r^{2}dr$$ if we bring spherical shells with width dr for example, but why would we integrate from $$0$$ to $$R$$ and not from $$R$$ to $$0$$? How do we decide that? I lack any physical reason for making the choice. It is the same with $$U=\frac{\epsilon_0}{2}\int E^2 dv$$ , I always see the summation going from radius 0 to infinity and not the reverse. Why?

• feynmanlectures.caltech.edu/II_08.html – Phoenix87 Apr 22 at 10:53
• Your formula is not valid for "some shape of charge". The charge Q must be spherically symmetrical. It is also not clear that you are adding the dq to Q. – R.W. Bird Apr 22 at 14:39
• I know, sorry, I wasn't accurate and quite confused while writing the question. – Darkenin Apr 22 at 15:26

Let's start from basic principles.

Works is defined as: $$dW=\vec F\cdot d\vec s\\\text{where}\space\vec F=k_e\frac{Qq}{r^2}\hat r$$ I have remembered online MIT video of deriving potential gravitational potential energy, which form I will use in derivation of our work. We are describing position in polar coordinates (we could do them in spherical coordinates, which would yield the same result, however complicate things).

From the image, we can see, that: $$d\vec s=rd\theta\hat \theta+dr\hat r$$ I.e. the differential displacement in polar coordinates has a component of r and theta. (In spherical coordinates we would have additional phi component).

Now here is the key fact. In polar/spherical coordinate system, the unit vectors are orthogonal. (See wikipedia for more info.)

We are finally ready to fully derive our work, without any ambiguities:

$$W_{total}=\int_{r_i}^{r_f}\vec F\cdot d\vec s=\int_{r_i}^{r_f}k_e\frac{Qq}{r^2}\hat r\cdot [rd\theta\hat \theta+dr\hat r]=\int_{r_i}^{r_f}k_e\frac{Qq}{r^2}dr\\\Biggl(\hat r\cdot\hat r = 1,\space\space \hat r \cdot \hat \theta = 0,\space(\hat r \cdot \hat \phi = 0)\Biggr)$$

Evaluating our integral, we obtain: $$W_{total}=-k_eQq\biggl[\frac{1}{r_f}-\frac{1}{r_i}\biggr]\\\text{In our case}\space\space r_i \to \infty\implies\frac{1}{r_i}\to0\\\text{We can also let }r_f\equiv r\\\text{to obtain}\space W(r)=-k_e\frac{Qq}{r}$$ We can see, that derived equation agrees with our intuition.

If we bring signed charges, they will 'repel' each other if same sign, thus requiring negative work, and 'attract' each other if different sign, doing work.

As an exercise, can you derive potential function to tell, when potential will be maximized and if it makes sense to have positive potential for different signed charges?

• You used the electric force as the force we are exerting when bringing the charge from infinity,but should't it be $F_{external} = -F_{electric}$ ? – Darkenin Apr 22 at 13:18
• Please check en.wikipedia.org/wiki/Coulomb%27s_law#The_law The sign can be replaced if we consider force acting on body 2 from body 1, instead of forces acting on body 1 from body 2 (By Newton's 3rd Law). I.e. in our vector form, we take position vector from Q to q. – Alexander Issa Apr 22 at 14:05
• Excuse me, but I'm still baffled by your answer. You said that if both charges are positive for example, the work done needs to be negative. How does that make sense? If there is a positive charge Q at some point in space and I do work to bring a positive test charge q to some distance from him, I need to exert force in the direction of the displacement (the q charge gets repelled and I push him toward Q). What don't I get right? I also see most derivations of this work without the minus you got there. – Darkenin Apr 22 at 14:59
• As you have correctly noted, the q charge gets repelled if brought towards Q with the same charge sign. It means, that as you displace the charge q towards Q, the force on q is opposite to that of displacement. I.e. the work turns negative by definition. Please give a thought to the exercise, and see what happens to potential, when the work described is negative? Also think about such scenario: Work can be described as change in kinetic energy. What happens if we 'throw' our charge q, with some velocity v_0 towards Q, until it stops if they have the same charge? How does it translate to work? – Alexander Issa Apr 22 at 15:29
• Correct me if I'm wrong, but you're referring to the force exerted by the electrostatic force, and I refer to the force the external agent does - because the force exerted by the pushing force itself (external) is in the direction of displacement in this case. – Darkenin Apr 22 at 16:03