# Work done by electromagnetic forces when considering continous charge distributions

Here’s something that has always bothered me in my physics lectures. Suppose I want to calculate the work done by electromagnetic forces onto one charge

$$$$dW = \vec{F} \cdot d\vec{s} = Q \vec{E}\cdot\vec{v} dt$$$$

Now if I want to expand this idea to a continuous charge distributions $$\rho$$ my textbook teels me to do the following using $$Q =\rho dV$$ and $$\vec{J}=\rho \vec{v}$$.

$$$$dW = \rho dV \vec{E}\cdot \frac{1}{\rho} \vec{J} dt = \vec{E}\cdot\vec{J} dV dt$$$$

Now up until here I am thinking about the values $$dW, dV, dt$$ as finite differences. Now making the switch to Infinitesimal values is what causes problems for me. In my textbook the next line simply states $$$$\frac{dW}{dt} = \int_{\Omega}^{} \vec{E}\cdot\vec{J} \,dV '$$$$

Now I have a couple of ussues here: Where do I suddenly pull out the integral? Why can i just divide by $$dt$$ as if it's a regular variable?

It's unclear why your integral at the end is not integrating over $$dV$$ (the name for the volume element $$dV$$ seems to be suddenly changed to $$d\Omega'$$, but I assume that is not the real question here).

Also, working with these separate "$$d$$" infinitesimals is a bit sloppy, mathematicians will probably tell you that you always need to combine them in some derivative of something with respect to something else, in a fraction like $$dA/dB$$.

But apart from that, you actually can can divide by $$dt$$ here, but that would not finish the job, because the "$$dW$$" in fact should have had two $$d$$'s, since you want it to be a finite difference in time and also in volume (there you see how things easily get messed up with this notation!)

So if you start with $$$$d(dW) = \vec{E}\cdot\vec{J} dV dt$$$$ and divide by $$dt$$ that would leave you with $$$$d\left(\frac{dW}{dt}\right) = \vec{E}\cdot\vec{J} dV$$$$ and applying an integral over space would remove one $$d$$ from the left-hand side, since the right-hand side is combining an integral with a $$d$$ which means summing over finite differences so it gives a total result: $$$$\frac{dW}{dt} = \int_{\Omega}^{} \vec{E}\cdot\vec{J} \,dV$$$$ At least that is the total result for $$dW/dt$$, not yet for $$W$$ because that would require another integral, over time (but to do that the division by $$dt$$ should not have been done, instead $$dt$$ should have remained on the right, and combined with a second integral: $$$$W = \int dt\ \int_{\Omega}^{} \vec{E}\cdot\vec{J} \,dV$$$$

So there are two games you can play with the $$d$$'s. Not only moving them around (e.g. by dividing) but also contracting them with $$\int$$ to create total values.

• ah sorry for the dV confusion, I fixed my typo. Mar 6 at 18:32
• What do these "d's" even mean? Also how does throwing an integral onto that suddenly fix it? Mar 7 at 8:18
• It's Leibniz's notation and people found it clearer than Newton's "fluxions", I think... And if an integral is summing a lot of small changes into a total result then you can forget about the small changes afterwards. (Perhaps not "suddenly", integrals can take time!) Mar 7 at 8:51